Month: November 2024

Jharkhand Board JAC Class 9 Maths Solutions Chapter 7 Triangles Ex 7.4 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.4

Page-132

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side

Answer:
ABC is a triangle right angled at B.
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90° (∵ ∠B = 90°)
So ∠A < 90° and ∠C < 90°
⇒ ∠B > ∠A and ∠B > ∠C
⇒ AC > BC and AC > AB [As side opposite to larger angle is longer]

Question 2.
In Fig, sides AB and AC of AABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Answer:
Given: ∠PBC < ∠QCB
Proof: ∠ABC + ∠PBC = 180°
⇒ ∠PBC = 180° – ∠ABC …(i)
Also, ∠ACB + ∠QCB = 180°
⇒ ∠QCB = 180° -∠ACB
⇒ ∠PBC < ∠QCB (Given)
⇒ 180° – ∠ABC < 180° – ∠ACB
(From (i) and (ii))
⇒ – ∠ABC < – ∠ACB
⇒ ∠ABC > ∠ACB
⇒ AC > AB (As sides opposite to larger angle is longer.)

Question 3.
In Fig, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Answer:
Given: ∠B < ∠A and ∠C < ∠D
Proof: In ΔAOB,
∠B < ∠A ⇒ AO < BO ……….(i)
(Side opposite to smaller angle is smaller)
In ΔCOD
∠C < ∠D ⇒ DO < CO…(ii)
(Side opposite to smaller angle is smaller)
Adding (i) and (ii)
AO + OD < BO + OC
⇒ AD < BC

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig). Show that ∠A > ∠C and ∠B > ∠D.

Answer:
In ΔABD,
AB < AD (As AB is the shortest side)
∴ ∠ADB < ∠ABD ……..(i) (Angle opposite to longer side is larger.)
Now, in ΔBCD,
BC < DC (As CD is the longest side)
∴ ∠BDC < ∠CBD …..(ii) (Angle opposite to longer side is larger.)
Adding (i) and (ii), we get
∠ADB + ∠BDC < ∠ABD + ∠CBD
⇒ ∠ADC < ∠ABC
⇒ ∠B > ∠D
Similarly, in Δ ABC,
AB < BC (As AB is the smallest side)
∠ACB < ∠B AC …(iii) (Angle opposite to longer side is larger)
Now, in ΔADC,
AD < DC (As CD is the longest side)
∠DCA < ∠DAC …(iv) (Angle opposite to longer side is larger)
Adding (iii) and (iv), we get
∠ACB + ∠DCA < ∠BAC + ∠DAC
⇒ ∠BCD < ∠BAD
⇒ ∠A > ∠C

Question 5.
In Fig, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ

Answer:
Given: PR > PQ and PS bisects ∠QPR
To prove: ∠PSR > ∠PSQ
Proof: ∠PQR > ∠PRQ …….(i)
(PR > PQ as angle opposite to longer side is larger)
∠QPS = ∠RPS …(ii) (PSbisects ∠QPR)
∠PSR = ∠PQR + ∠QPS ………(iii)
(exterior angle of a triangle equals the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS …(iv)
(exterior angle of a triangle equals the sum of opposite interior angles)
Adding (i) and (ii)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [from (iii) and (iv)]

Page-133

Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer:
Given: Let l is a line segment and B is a point lying on it. We draw a line segment AB perpendicular to 1. Let C be any other point on 1.
To prove: AB < AC
Proof: In ∆ABC,
∠B = 90°
Now, ∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90°
⇒ ∠C < 90° (∠B = 90°)
⇒ ∠C < ∠B
∴ ∠C must be acute angle or ∠C < ∠B
⇒ AB < AC (Side opposite to the larger angle is longer.)

Jharkhand Board JAC Class 10 Sanskrit Solutions रचना वाक्य निर्माणम् Questions and Answers, Notes Pdf.

JAC Board Class 10th Sanskrit Rachana वाक्य निर्माणम्

पद एवं उनके वाक्य प्रयोग :

पदाधारितवाक्यानां निर्माणम्

1. राजानं – सभासदः राजानं पश्यन्ति।
2. मन्त्रिणा राजा मन्त्रिणा सह वार्ता करोति।
3. विदुषे – राजा विदुषे ग्रन्थं ददाति।
4. राज्ञः – राज्ञः वचनं पालनीयम्।

सर्वनामशब्दाधारितवाक्यानाम् निर्माणम्

1. एषा – एषा मम भगिनी।
2. भवती – भवती अध्यापिका अस्ति।
3. यूयं – यूयं संस्कृतं पठत।
4. सः – सः बालकः अस्ति।

उपसर्गाधारितवाक्यानाम् निर्माणम्

1. अधिवसति – शिवः कैलासम् अधिवसति।
2. अपकरोति – दुर्जनः सदा अपकरोति।
3. प्रभवति – हिमालयात् गङ्गा प्रभवति।
4. संहरति – राजा शत्रून् संहरति।

अव्ययाधारितवाक्यानाम् निर्माणम्

1. सदा – बटुः सदा पठति।
2. समया – ग्रामं समया नदी वहति।
3. प्रभृतिः – अहं बाल्यकालात् प्रभृतिः अत्र वसामि।
4. इदानीं – इदानी वृष्टिः भवति।

हलन्तशब्दाधारितवाक्यानाम् निर्माणम्

1. श्रद्धावान् – श्रद्धावान् ज्ञानं लभते।
2. वाक् – वाक् मधुरा अस्ति।
3. सरित् – सरितः वहन्ति।
4. महत् – रोगिसेवा महत् कार्यं भवति।

अजन्तशब्दाधारितवाक्यानाम् निर्माणम्

1. विष्णुः – विश्वस्य पालकः विष्णुः।
2. लता – अने शोभते लता।
3. धेनुः – नन्दिनी वसिष्ठस्य धेनुः।
4. अश्रुः – बालकस्य नेत्राभ्याम् एकैकम् अश्रुः पतति।

कारकाधारितवाक्यानाम् निर्माणम्

1. धिक् – धिक् दुर्जनम्। (दृष्टि व्यक्ति को धिक्कार है।)
2. सह – पुत्रः पित्रा सह गच्छति। (पुत्र पिता के साथ जाता है।)
रमेशः मया सह पठति। (रमेश मेरे साथ पढ़ता है।)
बालकाः मित्रः सह क्रीडन्ति। (बालक मित्रों के साथ खेलते हैं।)
3. नमः – गुरवे नमः। (गुरु के लिए नमस्कार है।)
4. क्रुध् – दुर्जनः सज्जनाय क्रुध्यति। (दुष्ट व्यक्ति सज्जन पर क्रोध करता है।)
पिता पुत्राय क्रुध्यति। (पिता पुत्र पर क्रोध करता है।)

प्रत्ययाधारित वाक्य निर्माणम्

1. पठित्वा – छात्रः अत्र पठित्वा गच्छति। (छात्र यहाँ पढ़कर जाता है।)
2. गत्वा – छात्रः विद्यालयं गत्वा पठति। (छात्र विद्यालय जाकर पढ़ता है।)
3. आगत्य – रामः अत्र आगत्य पठति। (राम यहाँ आकर पढ़ता है।)
4. प्रणम्य – सेवकः प्रणम्य अवदत्। (सेवक प्रणाम करके बोला।)

पदाधारित वाक्य निर्माणम्

1. आसन् – बालकाः कक्षायाम् आसन्। (बालक कक्षा में थे।)
2. कुर्वन्ति – छात्राः गृहकार्यं कुर्वन्ति। (छात्र गृहकार्य करते हैं।)
3. अपतत् – सोहनः अश्वात् अपतत्। (सोहन घोड़े से गिर पड़ा।)
4. गच्छताम् – तौ गृहं गच्छताम्। (वे दोनों घर जावें।)

अभ्यासः

प्रश्न 1.
अधोलिखितशब्दानां सहायतया वाक्यनिर्माणं कुरुत-(निम्न शब्दों की सहायता से वाक्य-निर्माण कोजिए।)
[जीवितम्, शरणम्, दशनैः, निर्मलम्, करणीयम्, भृशम्, नगरात्, कान्तारे, क्षणमपि, कलरवः, जनेभ्यः, वनप्रदेशम्।]
उत्तरम् :
जीवितम् – प्रदूषितपर्यावरणे जीवितं कठिनं जातम्। (प्रदूषित पर्यावरण में जीवन कठिन हो गया है।)
शरणम् – श्रीकृष्णः शरणं मम। (श्रीकृष्ण मेरे आश्रय हैं।)
दशनैः – मानवः दशनैः भोजनं खादति। (मनुष्य दाँतों से खाना खाता है।)
निर्मलम् – निर्मलम् एव जलं पिबेत्। (शुद्ध जल ही पीना चाहिए।)
करणीयम् – कर्त्तव्यम् एव करणीयम्। (करने योग्य कार्य करना चाहिए।)
भृशम् – अद्य जलं भृशं दूषितम्। (आज पानी बहुत प्रदूषित है।)
नगरात – नगरात बहिरेव एकम् उद्यानम् अस्ति। (नगर के बाहर ही एक बाग है।)
कान्तारे – सीता निर्जनकान्तारे रामेण सह अगच्छत्। (सीता निर्जन वन में राम के साथ गई।)
क्षणमपि – न कश्चित् क्षणमपि तिष्ठति अकर्मकृत्। (कोई क्षणभर भी बिना काम किए नहीं रहता।)
कलरवः – खगानां कलरवः अतिमनोहरः भवति। (पक्षियों का कलरव मनोहर होता है।)
जनेभ्यः – स्वस्ति स्यात् जगतां जनेभ्यः। (संसार के लोगों का कल्याण हो।)
वनदेशम् – पशवः प्रातरेव वनप्रदेशं गच्छन्ति। (पशु प्रातः ही वन प्रदेश को जाते हैं।)

प्रश्न 2.
अधोलिखितानां पदानां सहायतया वाक्यनिर्माणं कुरुत-(निम्नलिखित शब्दों की सहायता से वाक्य बनाइए-)
[विमुक्ता, पितुहम्, दृष्ट्वा]
उत्तरम् :
विमुक्ता – सा व्याघ्रमारी निजबुद्ध्या व्याघ्रभयाद् विमुक्ता। (वह व्याघ्रमारी अपनी बुद्धि से बाघ के भय से मुक्त हो गई।)
पितुर्ग्रहम् – नार्यः रक्षाबन्धनोत्सवे पितुर्गृहं गच्छन्ति। (स्त्रियाँ रक्षाबन्धन के त्यौहार पर पिता के घर जाती हैं।)
दृष्ट्वा – सिंह दृष्ट्वा भयाद् असौ पलायितवान्। (सिंह को देखकर डर से वह भाग गया।)

प्रश्न 3.
अधोलिखितशब्दानां सहायतया वाक्यानां निर्माणं कुरुत- (निम्नलिखित शब्दों की सहायता से वाक्य का निर्माण कीजिए)-
[कृत्वा, सुखम्, समन्ततः, गात्राणाम्, लाघवम्, सुविभक्तता, आरोग्य, परमम्, पिपासा, सहसा]
उत्तरम् :
कृत्वा – सा गृहकार्यं कृत्वा विद्यालयं गच्छति। (वह गृहकार्य करके विद्यालय को जाती है।)
सुखम् – यावत् जीवेत् सुखं जीवेत्। (जब तक जीओ सुख से जीओ।)
समन्ततः – जनाः समन्ततः तत्र आगताः (लोग सभी ओर से वहाँ आ गये।)
गात्राणाम् – गात्राणाम् अनीशोऽहमस्मि। (मैं अपने अंगों का भी स्वामी नहीं हूँ।)
लाघवम् – मनुष्यः प्रयत्नलाघवम् अनुसरति। (मनुष्य प्रयत्न लाघव का अनुसरण करता है। )
सुविभक्तता – गात्राणाम् अपि सुविभक्तता शोभते। (अंगों का सुन्दर विभाजन शोभा देता है।)
आरोग्यम् – आरोग्यं परमं सुखम्। (नीरोग होना सबसे बड़ा सुख है।)
परमम् – दारिद्रयं परमापदां पदम्। (दरिद्रता महान आपदाओं का स्थान है।)
पिपासा – ग्रीष्मे पिपासा अतिबाधते। (गर्मियों में प्यास अधिक कष्ट देती है।)
सहसा – सहसा विदधीत न क्रियाम्। (अकस्मात् कोई काम नहीं करना चाहिए।)

प्रश्न 4.
अधोलिखितशब्दानां सहायतया वाक्यनिर्माणं कुरुत – (निम्नलिखित शब्दों की सहायता से वाक्य-निमाण कीजिए-)
[उपसृत्य, अलम्, लालनीयः, खलु, नामधेयम्, धिक, अपूर्वः, श्लाघ्या, त्वरयति।]
उत्तरम् :
उपसृत्य – रामः जनकम् उपसृत्य प्रणमति। (राम पिता के पास जाकर प्रणाम करता है।)
अलम् – अलं विवादेन। (विवाद मत करो।)
लालनीयः – शिशु पित्रोः लालनीयः भवति। (बच्चा पिता का लाडला होता है।)
खलु – मा खलु चापलं कुरु। (निश्चित ही चपलता मत करो !)
नामधेयम् – किं ते नामधेयम् ? (तेरा क्या नाम है ?)
धिक् – धिक् त्वाम्। (तुझे धिक्कार है।)
अपूर्वः – अपूर्वः कोऽपि कोशोऽयं विद्यते तव भारती। (हे सरस्वती ! यह तुम्हारा खजाना अनोखा है।)
श्लाघ्या – श्लाघ्या इयं रामायणी कथा। (यह रामायण की कथा सराहनीय है।)
त्वरयति – श्रमस्य फलं मां त्वरयति। (श्रम का फल मुझसे जल्दी करवा रहा है।)

प्रश्न 5.
अधोलिखित पदानां सहायताया वाक्य निर्माणं कुरुत। (निम्न पदों से वाक्य बनाइये)
[हलमा पपात, जवन, कषावल, वषभ, भारम, धेनूनाम्, अश्रूणि, वासवः, बोदुम्।]
उत्तरम् :
हलम् – वृषभौ हलम् स्कन्धो धृत्वा क्षेत्रं गच्छतः। (बैल हल लेकर खेत पर जाते हैं।)
पपात – एकं पाषाण खण्डं उपरितः पपात तेन सः हतः। (एक पत्थर का टुकड़ा ऊपर से गिरा, जिससे वह मर गया।)
जवेन – सः जवेन गच्छति एव, मार्गेऽहं मिलितवान्। (वह तेज गति से जा ही रहा था कि रास्ते में मैं मिल गया)
कृषीवलः – कृषीवल: अन्नमुत्पादयति। (किसान अन्न पैदा करता है।)
वृषभः – वृषभः कृषिकार्ये कृषकस्य सदैव सहायतां करोति। (बैल खेती के काम में किसान की सदैव सहायता करता है।)
भारम् – निर्बलः जनः भारम् न सोढुं शक्नोति। (निर्बल व्यक्ति भार नहीं ढो सकता।)
धेनूनाम् – सर्वासां धेनूनाम् जननी सुरभिः। (सभी गायों की माता सुरभि है।)
अश्रूणि – तस्य नेत्राभ्याम् अश्रूणि पतन्ति। (उसकी आँखों से आँसू गिरते हैं।)
वासवः – इन्द्रः वासवः इति नाम्नानि ख्यातः। (इन्द्र वासव के नाम से भी प्रसिद्ध हैं।)
बोढुम् – स्वस्थ मनुष्य एव भारं बोढुं शक्नोति। (स्वस्थ मनुष्य ही बोझा ढो सकता है।)

प्रश्न 6.
अधोलिखितपदानां सहायतया वाक्यनिर्माणं कुरुत (निम्नलिखित पदों की सहायता से वाक्य बनाइए)
उत्तरम् :
महान् – सुभाषः महान् पुरुषः आसीत्। (सुभाष महान् पुरुष था।)
समः – न कोऽपि वीरः रामेण समः। (राम के समान कोई वीर नहीं।)
कृत्वा – बालकः गृहकार्यं कृत्वा विद्यालयं गच्छति। (बालक गृहकार्य करके विद्यालय जाता है।)
ध्रुवम् – ध्रुवम् एव सः आयास्यति। (निश्चित ही वह आयेगा।)
अपगमे – मेघजालस्यापगमे चन्द्रः दृश्यते। (मेघजाल के हट जाने पर चन्द्रमा दिखाई पड़ता है।)
निमित्तम् – मूर्खाः किमपि निमित्तं विधाय कलहं कुर्वन्ति। (मूर्ख कोई कारण रखकर झगड़ा करते हैं।)
प्रथमः – पुस्तके प्रथमः पाठः मङ्गलाचरणं भवति। (पुस्तक में पहला पाठ मंगलाचरण होता है।)
यथा – यथा वृष्टिः भवति तथैव जलं प्रवहति। (जैसे वर्षा होती है वैसे ही जल बहता है।)
एव – माम् सः एव जानाति। (मुझे वह ही जानता है।)
सेवितव्यः – आचार्यः सर्वतोभावेन सेवितव्यः। (आचार्य की पूर्ण रूप से सेवा करनी चाहिए।)

प्रश्न 7.
अधोलिखित पदानां सहायतया वाक्य निर्माणं कुरुत – (निम्न पदों की सहायता से वाक्य निर्माण कीजिए।)
[समीपे, प्रहर्तुम्, आरूढः, आकृष्य, तुदन्ति, गर्जति, इतस्ततः, सन्नपि, जायते, किमर्थम्।]
उत्तरम् :
समीपे – समीपे एव उद्यानमस्ति यत्र वयं भ्रमामः। (पास में एक उद्यान है जहाँ हम घूमते हैं।)
प्रहर्तुम् – चौरः गृहस्वामिनः उपरि प्रहर्तुम् ऐच्छत। (चोर ने गृहस्वामी पर प्रहार करना चाहा।)
आरूढः – रथम् आरूढ़ः अर्जुनः शत्रून हन्ति। (रथ पर चढ़ता हुआ अर्जुन शत्रुओं को मारता है।)
आकृष्य – वानरः सिंहस्य कर्णमाकृष्य वृक्षमारोहति। (वानर सिंह का कान खींचकर पेड़ पर चढ़ जाता है।)
तुदन्ति – किमर्थं मामेव तुदन्ति सर्वे मिलित्वा। (तुम सभी मिलकर क्यों तंग कर रहे हो?)
गर्जति – सिंहः उच्चैः गर्जति। (सिंह जोर से दहाड़ता है।)
इतस्ततः – वानरं दृष्ट्वा बालकः इतस्ततः धावन्ति। (बन्दर को देखकर बालक इधर-उधर दौड़ते हैं।)
सन्नपि – सिंह सन्नपि वानरेभ्य विभेति। (शेर होते हुए भी वानरों से डरते हो।)
जायते – कामात् जायते क्रोधः। (काम से क्रोध पैदा होता है।)
किमर्थम् – त्वं किमर्थम् अत्र आगतोऽसि? (तुम यहाँ किसलिए आये हो?)

प्रश्न 8.
अधोलिखितानां पदानां सहायतया वाक्यनिर्माणं कुरुत – (निम्नलिखित पदों की सहायता से वाक्य बनाइये।)
उत्तरम् :
भूरि – सः निर्धनेभ्यः भूरि धनमयच्छत्। (उसने निर्धनों के लिए बहुत-सा धन दिया।)
संलग्नः – साधुः स्वकार्ये संलग्नः अभवत् (साधु अपने काम में लग गया।)
विचार्य – किञ्चित् विचार्य सोऽवदत्। (कछ विचार कर वह बोला।)
उपस्थातुम् – अंहं विद्यालये उपस्थातुम् असमर्थोऽस्मि। (मैं विद्यालय में उपस्थित होने में असमर्थ हूँ।)
अन्येधुः – अन्येधुः सः विदेशं गतवान्। (दूसरे दिन वह विदेश चला गया।)
निकषा – ग्रामं निकषा पाठशाला अस्ति। (गाँव के पास पाठशाला है।)।
वक्तुम् – अपि किमपि वक्तुम् इच्छति भवान्। (क्या आप कुछ कहना चाहते हैं ?)
अध्वनि – स: मां अध्वनि सर्वम् एव वृत्तम् अकथयत्। (उसने मुझसे मार्ग में सारा वृत्तान्त कह दिया।)
भुक्ष्व – भुक्ष्व इदानी स्वकर्मणः फलम्। (भोग अब अपने कर्म का फल।)

प्रश्न 9.
अधोलिखित शब्दानां सहायतया वाक्य निर्माणं कुरुत –
(निम्नलिखित शब्दों की सहायता से वाक्य निर्माण कीजिये।)
[यच्छति, विद्याधनम्, चित्ते, वाचि, त्यक्त्वा, पक्वं, वदने, प्रोक्तम्, परिभूयते, आत्मनः]
उत्तरम् :
यच्छति – धनिकः भिक्षुकेभ्यः भोजनं यच्छति। (धनवान भिक्षुओं को भोजन देता है।)
विद्याधनम् – विद्याधनं सर्वधन प्रधानम्। (विद्याधन सभी धनों में प्रमुख है।)
चित्ते – न चित्ते केभ्यः अपि अहितभावः भवेत्। (मन में किसी के लिये अहित का भाव न हो।)
वाचि – वाचि सत्यम् भवेत्। (वाणी में सत्यता होनी चाहिये।)
त्यक्त्वा – परपीडनं त्यक्त्वा सदैव परोपकारमेव कुर्यात्। (दूसरों को दुख देने को त्यागकर परोपकार करना चाहिये।)
पक्वं – पक्वमेव फलं मधुरं भवति। (पका हुआ फल ही मीठा होता है।)
वदने – तस्य वदने स्मिति तस्य प्रसन्नतां दर्शयिति। (उसके मुँह पर मुस्कराहट उसकी प्रसन्नता को दर्शाती है।)
प्रोक्तम् – इति महापुरुषैः प्रोक्तम्। (ऐसा महापुरुषों ने कहा)
परिभूयते – मूर्खः सर्वत्रैव परिभूयते। (मूर्ख सब जगह अपमानित किया जाता है।)
आत्मनः – आत्मनः हितचिन्तनमेव स्वार्थम्। (अपना हित सोचना ही स्वार्थ है।)

प्रश्न 10.
अधोलिखितानां पदानां सहायतया वाक्यनिर्माणं कुरुत।
(निम्नलिखित पदों की सहायता से वाक्य-निर्माण कीजिए।)
[जातम्, अकस्मात्, बहुभूमिकानि, महत्कम्पनम्, भीषणम्, विभीषिका, खलु, वस्तुतः, निर्माय।]
उत्तरम् :
जातम् – स्वातन्त्र्यं प्राप्य भारतं राष्ट्र प्रसन्नं जातम्। (स्वतन्त्रता प्राप्त करके भारत राष्ट्र प्रसन्न हो गया।)
अकस्मात् – अकस्मात् एव युद्धस्य स्थितिः उपस्थिता। (अचानक ही युद्ध की स्थिति उपस्थित हो गई।)
बहुभूमिकानि – महानगरेषु बहुभूमिकानि भवनानि सन्ति। (महानगरों में बहुमंजिले भवन हैं।)
महत्कम्पनम् – भूकम्पे जाते धरायाः महत् कम्पनम् आरभते। (भूकम्प आने पर धरती का अत्यन्त काँपना आरम्भ होता है।)
भीषणम् – कौरवपाण्डवयोः भीषणं युद्धम् अजायत। (कौरव-पाण्डवों का भीषण युद्ध हुआ।)
विभीषिका – भूकम्पस्य विभीषिका युद्धाद् अपि भयङ्करा। (भूकम्प की आपदा युद्ध से भी अधिक भयंकर है।)
खलु – मा खलु चापलं कुरु। (निश्चय ही चपलता मत करो।)
वस्तुतः – वस्तुतः प्रदूषणम् एव रोगाणां कारणम्। (वास्तव में प्रदूषण ही रोगों का कारण है।)
निर्माय – खगाः अपि नीडं निर्माय निवसन्ति। (पक्षी भी घोंसला बनाकर निवास करते हैं।)

प्रश्न 11.
अधोलिखितानां पदानां सहायतया वाक्य-निर्माणं कुरुत – (निम्नलिखित शब्दों की सहायता से वाक्य बनाइये)
[द्रष्टुम्, निष्क्रम्य, प्रविश्य, पिधाय, भवान्, रक्षसि, क्व, ज्ञायते, दुष्करम्।]
उत्तरम् :
द्रष्टुम् – बालकः चलचित्रं द्रष्टुम् इच्छति। (बालक चलचित्र देखना चाहता है।)
निष्क्रम्य – मूषक: बिलात् निष्क्रम्य भूमौ लुण्ठति। (चूहा बिल से निकलकर धरती पर लोटता है।)
प्रविश्य – सा कक्षां प्रविश्य आचार्यां नमति। (वह कक्षा में प्रवेश करके आचार्या को नमस्कार करती है।)
पिधाय – धनिकः कुम्भं पिधाय भूमौ निक्षिप्तवान्। (धनवान् ने घड़े को ढंककर धरती में रख दिया।)
भवान् – भवान् कुत्र गच्छति ? (आप कहाँ जा रहे हैं ?)
रक्षसि – हे ईश्वर ! त्वं सर्वान् रक्षसि। (हे ईश्वर ! तू सबकी रक्षा करता है।)
क्व – इदानीं सः धूर्तः क्व गतः ? (अब वह चालाक कहाँ गया ?)
ज्ञायते – सर्वमिदं मया ज्ञायते। (यह सब मुझे ज्ञात है।)
दुष्करम् – कः कुर्यात् इदं दुष्करं शिविना विना ? (शिवि के बिना इस दुष्कर कार्य को और कौन करे!)

प्रश्न 12.
अधोलिखितानां शब्दानां सहायतया वाक्यनिर्माणं कुरुत – (निम्नलिखित शब्दों की सहायता से वाक्य बनाइये।)
उत्तरम् :
भुक्तम् – तेन भोजनं भुक्तम्। (उसने खाना खाया।)
निपीतम् – मया भोजनान्ते तक्रं निपीतम्। (मैंने भोजन के बाद छाछ पी है।)
वद – सत्यं वद। (सत्य बोल।)
परितः – ग्रामं परितः राजमार्गः। (गाँव के चारों ओर सड़क है।)
रसालमुकुलानि – भ्रमराः वसन्तकाले रसालमुकुलानि आश्रयन्ते। (भौरे वसन्तकाल में आम के बौर का आश्रय लेते हैं।)
हन्त – हन्त ! अनेके वीराः हताः युद्धे। (खेद है, अनेक वीर युद्ध में मारे गये।)
पूरयित्वा – वारिदः जलाशयान् पूरयित्वा रिक्तः भवति। (बादल जलाशयों को भरकर खाली हो जाता है।)
काननानि – सीता लक्ष्मणेन सह काननानि द्रष्टुम् अगच्छत्। (सीता लक्ष्मण के साथ जंगलों को देखने गई।)
जलदः – जलदः वर्षति जलं च प्रवहति। (बादल बरसता है और पानी बहता है।)

Jharkhand Board JAC Class 9 Maths Solutions Chapter 8 Quadrilaterals Ex 8.1 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

Page-146

Question 1.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Answer:
Let the angles be 3x, 5x, 9x and 13x.
Sum of the interior angles of the quadrilateral = 360°
Now, 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°
⇒ x = 12°
Angles of the quadrilateral are:
3x = 3 × 12° = 36°
5x = 5 × 12° = 60°
9x = 9 × 12° = 108°
13x= 13 × 12°= 156°

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:
Given: AC = BD
To prove: ABCD is a rectangle.
Proof: In order to prove ABCD is a rectangle we need to prove that one of its interior angle is right angle.
In ΔABD and ΔBAD,
BA = BA (Common)
BC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC = ΔBAD by SSS congruence criterion.
⇒ ∠A = ∠B (by CPCT)
Also, ∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Thus, ABCD is a rectangle.

Question 3.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given: OA = OC, OB = OD and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
To prove: ABCD is parallelogram and AB = BC = CD = AD
Proof: In ΔAOB and ΔCOB,
OA = OC
∠AOB = ∠COB = 90°
OB = OB (Common)
Therefore, ΔAOB ≅ ΔBOC by SAS congruence criterion.
Thus, AB = BC (by CPCT)
Similarly, we can prove,
AB = BC = CD = AD Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus.

Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.

Answer:
Given: Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To prove: AC = BD, AO = OC and ∠AOB = 90°
Proof: In ΔABC and ΔBAD,
AB = BA (Common)
∠ABC = ∠BAD = 90° (Each angle of square is of 90°)
BC = AD (All sides of square are equal)
Therefore, ΔABC ≅ ΔBAD by SAS congruence criterion.
Thus, AC = BD (by CPCT).
i.e., diagonals are equal.

Now, In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite angles)
AB = CD (Given)
Therefore, ΔAOB = ΔCOD by AAS congruence criterion.
Thus, AO = CO and BO = DO by CPCT i.e.
Diagonals bisect each other.
Now, In ΔAOB and ΔCOB,
OB = OB (common)
AO = CO (Proved above)
AB = CB (Sides of the square)
Therefore, ΔAOB ≅ ΔCOB by SSS congruence criterion.
∴ ∠AOB = ∠COB (CPCT)
Also, ∠AOB + ∠COB = 180°
(Linear pair)
Thus, ∠AOB = ∠COB = 90° i.e.
Diagonals bisect each other at right angles.

Question 5.
Show that if the diagonals of a which diagonals AC and BD bisect each other at right angles at O.

Answer:
Given: Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angles at O.
To prove: Quadrilateral ABCD is a square.
Proof: In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite angles)
OB = OD (Diagonals bisect each other)
Therefore, ΔAOB ≅ ΔCOD by SAS congruence criterion.
Thus, AB = CD by CPCT. ……. (i)
Now, In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD = 90° (Diagonals bisect each other at right angle)
OD = OD (Common)
Therefore, ΔAOD ≅ ΔCOD by SAS congruence criterion.
Thus, AD = CD by CPCT
Similarly, ΔCOD ≅ ΔBOC and BC = CD (CPCT)
⇒ AD = BC = CD = AB ……… (ii)
Also, as diagonals bisect each other, so, ABCD is a parallelogram.

Consider ΔACD and ΔBDC
AD = BC (Proved above)
CD = CD (Common)
AC = BD (Given)
⇒ ΔACD ≅ ΔBDC (by SSS congruence criterion)
⇒ ∠ADC = ∠BCD (CPCT)
As ABCD is a parallelogram so,
AD || BC
⇒ ∠ADC + ∠BDC = 180° (Angles on the same side of transversal)
∴ ∠ADC – ∠BCD = 90°.
So, ABCD is a parallelogram in which all sides are equal and one angle is 90°
⇒ ABCD is a square.

Question 6.
Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig.). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.

Answer:
(i) As ABCD is a parallelogram
⇒ AB || CD and AD || BC
⇒ ∠BAC = ∠ACD …….(i)
(Alternate interior angles)
∠CAD = ∠ACB ……. (ii)
(Alternate interior angles)
Also, ∠CAD = ABAC …….. (iii)
(As AC bisects ∠A)
Therefore, ∠BCA=∠DCA [By (i), (ii), (iii)]
⇒ AC bisects ∠C.

(ii) As ABCD is a parallelogram
⇒ AD = BC and AB = CD …(iv)
Also, as ∠CAD = ∠ACB (Alternate interior angles)
and ∠ACB = ∠ACD (Proved)
⇒ ∠CAD = ∠ACD
⇒ AD = CD …(v)
(sides opposites to equal angles are equal)
Similarly, AB = BC …(vi)
Therefore, AB = BC = CD = AD (By (iv), (v) (vi))
So, ABCD is a parallelogram in which all sides are equal.
⇒ ABCD is a rhombus.

Question 7.
ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and

Answer:
Let ABCD is a rhombus and AC and BD be its diagonals.
Proof: In ΔACD – AD = CD (Sides of a rhombus)
∠DAC = ∠DCA
(Angles opposite to equal sides of a triangle are equal)
Also, AB || CD (Opposite sides of rhombus are parallel)
⇒ ∠DAC = ∠BCA (Alternate interior angles)
⇒ ∠DCA = ∠BCA
Therefore, AC bisects ∠C.
In ∆ABC,
AB = BC (Sides of a rhombus)
⇒ ∠BAC = ∠BCA (Angles opposite to equal sides are equal)
Also, AD || BC (Opposite sides of rhombus are parallel)
⇒ ∠DAC = ∠BCA (Alternate interior angles)
∴ ∠BAC = ∠DAC
⇒ AC bisects ∠A
Similarly we can prove that diagonal BD bisects ∠B as well as ∠D.

Question 8.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.

Answer:
(i) AB || CD and AD || BC (∵ opposite sides of rectangle are parallel)
⇒ ∠1 = ∠4 and∠2=∠3 (Alternate interior angles)
Also, as AC bisects ∠A and ∠C
⇒ ∠1 = ∠2 and ∠3 = ∠4
Therefore ∠2 = ∠4
So, in ΔACD,
AD = CD (sides opposite to equal angles are equal)
Also, AD = BC and AB = CD (∵ opposite sides of rectangle are equal)
∴ AB = BC = CD = AD
⇒ ABCD is a rectangle in which all sides are equal
⇒ ABCD is a square.

(ii) In ΔABD,
AB = AD (sides of a square)
⇒ ∠5 = ∠7 (Angles opposite to equal sides are equal)
Also, as AB || CD
⇒ ∠5 = ∠8 (Alternate interior angles)
∴ ∠7 = ∠8
⇒ BD bisects ∠D
Similarly, BD bisects ∠B.

Page-147

Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig). Show that:

(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Answer:
(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a ||^gm)
Thus, ΔAPD ≅ ΔCQB
(by SAS congruence criterion)

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = CD (Opposite sides of a ||^gm)
Thus, ΔAQB ≅ ΔCPD by SAS congruence criterion.

(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.

(v) From (ii) and (iv), we get opposite sides of quadrilateral APCQ are equal. Thus, APCQ is a ||^gm.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ

Answer:
(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD = 90° (∵ AP and CQ are perpendiculars)
AB = CD (Opposite sides of a is a parallelogram)
Thus, ΔAPB ≅ ΔCQD (by AAS congruence criterion)

(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.

Question 11.
In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.

Answer:
(i) AB = DE and AB || DE (Given)
Thus, quadrilateral ABED is a parallelogram because a pair of opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again BC = EF and BC || EF.
Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.
⇒ AD = BE and BE = CF
(Opposite sides of a parallelogram are equal)
Thus, AD = CF.
Also, AD || BE and BE || CF
(Opposite sides of a parallelogram are parallel)
Thus, AD || CF

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other.
Thus, ACFD is a parallelogram.

(v) AC || DF and AC = DF because ACFD is a parallelogram.

(vi) In ∆ABC and ∆DEF,
AB = DE (Given)
BC = EF (Given)
AC = DF (Opposite sides of a parallelogram ACFD)
Thus, ∆ABC ≅ ∆DEF (by SSS congruence criterion)

Question 12.
ABCD is a trapezium in which AB || CD and AD = BC.
Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Construction: Draw a line through C parallel to DA intersecting AB produced at E.
Proof:
(i) In quadrilateral ∆DCE
AE || CD (as AB || CD)
AD || CE (By construction)
∴ ∆DCE is a parallelogram as opposite sides are parallel
CE = AD (Opposite sides of a parallelogram)
AD = BC (Given)
Therefore, BC = CE
⇒ ∠CBE = ∠CEB
(Angles opposite to equal sides)
also, ∠A + ∠CBE = 180°
(Angles on the same side of transversal and ∠CBE = ∠CEB)
∠B + ∠CBE = 180° (Linear pair)
⇒ ∠A = ∠B

(ii) ∠A + ∠D = ∠B+ ∠C= 180° (Angles on the same side of transversal)
⇒ ∠A + ∠D = ∠A + ∠C (∵ ∠A = ∠B)
⇒ ∠D = ∠C

(iii) In ∆ABC and ∆BAD,
AB = AB (Common)
∠DAB = ∠CBA (Proved)
AD = BC (Given)
Thus, ∆ABC ≅ ∆BAD by SAS congruence criterion.

(iv) Diagonal AC = diagonal BD by CPCT as ∆ABC ≅ ∆BAD.

JAC Jharkhand Board Class 12th Maths Solutions in Hindi & English Medium

JAC Board Class 12th Maths Solutions in English Medium

JAC Class 12 Maths Chapter 1 Relations and Functions

• Chapter 1 Relations and Functions Ex 1.1
• Chapter 1 Relations and Functions Ex 1.2
• Chapter 1 Relations and Functions Ex 1.3
• Chapter 1 Relations and Functions Ex 1.4
• Chapter 1 Relations and Functions Miscellaneous Exercise

JAC Class 12 Maths Chapter 2 Inverse Trigonometric Functions

• Chapter 2 Inverse Trigonometric Functions Ex 2.1
• Chapter 2 Inverse Trigonometric Functions Ex 2.2
• Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

JAC Class 12 Maths Chapter 3 Matrices

• Chapter 3 Matrices Ex 3.1
• Chapter 3 Matrices Ex 3.2
• Chapter 3 Matrices Ex 3.3
• Chapter 3 Matrices Ex 3.4
• Chapter 3 Matrices Miscellaneous Exercise

JAC Class 12 Maths Chapter 4 Determinants

• Chapter 4 Determinants Ex 4.1
• Chapter 4 Determinants Ex 4.2
• Chapter 4 Determinants Ex 4.3
• Chapter 4 Determinants Ex 4.4
• Chapter 4 Determinants Ex 4.5
• Chapter 4 Determinants Ex 4.6
• Chapter 4 Determinants Miscellaneous Exercise

JAC Class 12 Maths Chapter 5 Continuity and Differentiability

• Chapter 5 Continuity and Differentiability Ex 5.1
• Chapter 5 Continuity and Differentiability Ex 5.2
• Chapter 5 Continuity and Differentiability Ex 5.3
• Chapter 5 Continuity and Differentiability Ex 5.4
• Chapter 5 Continuity and Differentiability Ex 5.5
• Chapter 5 Continuity and Differentiability Ex 5.6
• Chapter 5 Continuity and Differentiability Ex 5.7
• Chapter 5 Continuity and Differentiability Ex 5.8
• Chapter 5 Continuity and Differentiability Miscellaneous Exercise

JAC Class 12 Maths Chapter 6 Application of Derivatives

• Chapter 6 Application of Derivatives Ex 6.1
• Chapter 6 Application of Derivatives Ex 6.2
• Chapter 6 Application of Derivatives Ex 6.3
• Chapter 6 Application of Derivatives Ex 6.4
• Chapter 6 Application of Derivatives Ex 6.5
• Chapter 6 Application of Derivatives Miscellaneous Exercise

JAC Class 12 Maths Chapter 7 Integrals

• Chapter 7 Integrals Ex 7.1
• Chapter 7 Integrals Ex 7.2
• Chapter 7 Integrals Ex 7.3
• Chapter 7 Integrals Ex 7.4
• Chapter 7 Integrals Ex 7.5
• Chapter 7 Integrals Ex 7.6
• Chapter 7 Integrals Ex 7.7
• Chapter 7 Integrals Ex 7.8
• Chapter 7 Integrals Ex 7.9
• Chapter 7 Integrals Ex 7.10
• Chapter 7 Integrals Ex 7.11
• Chapter 7 Integrals Miscellaneous Exercise

JAC Class 12 Maths Chapter 8 Application of Integrals

• Chapter 8 Application of Integrals Ex 8.1
• Chapter 8 Application of Integrals Ex 8.2
• Chapter 8 Application of Integrals Miscellaneous Exercise

JAC Class 12 Maths Chapter 9 Differential Equations

• Chapter 9 Differential Equations Ex 9.1
• Chapter 9 Differential Equations Ex 9.2
• Chapter 9 Differential Equations Ex 9.3
• Chapter 9 Differential Equations Ex 9.4
• Chapter 9 Differential Equations Ex 9.5
• Chapter 9 Differential Equations Ex 9.6
• Chapter 9 Differential Equations Miscellaneous Exercise

JAC Class 12 Maths Chapter 10 Vector Algebra

• Chapter 10 Vector Algebra Ex 10.1
• Chapter 10 Vector Algebra Ex 10.2
• Chapter 10 Vector Algebra Ex 10.3
• Chapter 10 Vector Algebra Ex 10.4
• Chapter 10 Vector Algebra Miscellaneous Exercise

JAC Class 12 Maths Chapter 11 Three Dimensional Geometry

• Chapter 11 Three Dimensional Geometry Ex 11.1
• Chapter 11 Three Dimensional Geometry Ex 11.2
• Chapter 11 Three Dimensional Geometry Ex 11.3
• Chapter 11 Three Dimensional Geometry Miscellaneous Exercise

JAC Class 12 Maths Chapter 12 Linear Programming

• Chapter 12 Linear Programming Ex 12.1
• Chapter 12 Linear Programming Ex 12.2
• Chapter 12 Linear Programming Miscellaneous Exercise

JAC Class 12 Maths Chapter 13 Probability

• Chapter 13 Probability Ex 13.1
• Chapter 13 Probability Ex 13.2
• Chapter 13 Probability Ex 13.3
• Chapter 13 Probability Ex 13.4
• Chapter 13 Probability Ex 13.5
• Chapter 13 Probability Miscellaneous Exercise

JAC Board Class 12th Maths Solutions in Hindi Medium

JAC Class 12 Maths Chapter 1 संबंध एवं फलन

• Chapter 1 संबंध एवं फलन Ex 1.1
• Chapter 1 संबंध एवं फलन Ex 1.2
• Chapter 1 संबंध एवं फलन Ex 1.3
• Chapter 1 संबंध एवं फलन Ex 1.4
• Chapter 1 संबंध एवं फलन विविध प्रश्नावली

JAC Class 12 Maths Chapter 2 प्रतिलोम त्रिकोणमितीय फलन

• Chapter 2 प्रतिलोम त्रिकोणमितीय फलन Ex 2.1
• Chapter 2 प्रतिलोम त्रिकोणमितीय फलन Ex 2.2
• Chapter 2 प्रतिलोम त्रिकोणमितीय फलन विविध प्रश्नावली

JAC Class 12 Maths Chapter 3 आव्यूह

• Chapter 3 आव्यूह Ex 3.1
• Chapter 3 आव्यूह Ex 3.2
• Chapter 3 आव्यूह Ex 3.3
• Chapter 3 आव्यूह Ex 3.4
• Chapter 3 आव्यूह विविध प्रश्नावली

JAC Class 12 Maths Chapter 4 सारणिक

• Chapter 4 सारणिक Ex 4.1
• Chapter 4 सारणिक Ex 4.2
• Chapter 4 सारणिक Ex 4.3
• Chapter 4 सारणिक Ex 4.4
• Chapter 4 सारणिक Ex 4.5
• Chapter 4 सारणिक Ex 4.6
• Chapter 4 सारणिक विविध प्रश्नावली

JAC Class 12 Maths Chapter 5 सांतत्य तथा अवकलनीयता

• Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.1
• Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.2
• Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.3
• Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.4
• Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.5
• Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.6
• Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.7
• Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.8
• Chapter 5 सांतत्य तथा अवकलनीयता विविध प्रश्नावली

JAC Class 12 Maths Chapter 6 अवकलज के अनुप्रयोग

• Chapter 6 अवकलज के अनुप्रयोग Ex 6.1
• Chapter 6 अवकलज के अनुप्रयोग Ex 6.2
• Chapter 6 अवकलज के अनुप्रयोग Ex 6.3
• Chapter 6 अवकलज के अनुप्रयोग Ex 6.4
• Chapter 6 अवकलज के अनुप्रयोग Ex 6.5
• Chapter 6 अवकलज के अनुप्रयोग विविध प्रश्नावली

JAC Class 12 Maths Chapter 7 समाकलन

• Chapter 7 समाकलन Ex 7.1
• Chapter 7 समाकलन Ex 7.2
• Chapter 7 समाकलन Ex 7.3
• Chapter 7 समाकलन Ex 7.4
• Chapter 7 समाकलन Ex 7.5
• Chapter 7 समाकलन Ex 7.6
• Chapter 7 समाकलन Ex 7.7
• Chapter 7 समाकलन Ex 7.8
• Chapter 7 समाकलन Ex 7.9
• Chapter 7 समाकलन Ex 7.10
• Chapter 7 समाकलन Ex 7.11
• Chapter 7 समाकलन विविध प्रश्नावली

JAC Class 12 Maths Chapter 8 समाकलनों के अनुप्रयोग

• Chapter 8 समाकलनों के अनुप्रयोग Ex 8.1
• Chapter 8 समाकलनों के अनुप्रयोग Ex 8.2
• Chapter 8 समाकलनों के अनुप्रयोग विविध प्रश्नावली

JAC Class 12 Maths Chapter 9 अवकल समीकरण

• Chapter 9 अवकल समीकरण Ex 9.1
• Chapter 9 अवकल समीकरण Ex 9.2
• Chapter 9 अवकल समीकरण Ex 9.3
• Chapter 9 अवकल समीकरण Ex 9.4
• Chapter 9 अवकल समीकरण Ex 9.5
• Chapter 9 अवकल समीकरण Ex 9.6
• Chapter 9 अवकल समीकरण विविध प्रश्नावली

JAC Class 12 Maths Chapter 10 सदिश बीजगणित

• Chapter 10 सदिश बीजगणित Ex 10.1
• Chapter 10 सदिश बीजगणित Ex 10.2
• Chapter 10 सदिश बीजगणित Ex 10.3
• Chapter 10 सदिश बीजगणित Ex 10.4
• Chapter 10 सदिश बीजगणित विविध प्रश्नावली

JAC Class 12 Maths Chapter 11 त्रि-विमीय ज्यामिति

• Chapter 11 त्रिविमीय ज्यामिति Ex 11.1
• Chapter 11 त्रिविमीय ज्यामिति Ex 11.2
• Chapter 11 त्रिविमीय ज्यामिति Ex 11.3
• Chapter 11 त्रिविमीय ज्यामिति विविध प्रश्नावली

JAC Class 12 Maths Chapter 12 रैखिक प्रोग्रामन

• Chapter 12 रैखिक प्रोग्रामन Ex 12.1
• Chapter 12 रैखिक प्रोग्रामन Ex 12.2
• Chapter 12 रैखिक प्रोग्रामन विविध प्रश्नावली

JAC Class 12 Maths Chapter 13 प्रायिकता

• Chapter 13 प्रायिकता Ex 13.1
• Chapter 13 प्रायिकता Ex 13.2
• Chapter 13 प्रायिकता Ex 13.3
• Chapter 13 प्रायिकता Ex 13.4
• Chapter 13 प्रायिकता Ex 13.5
• Chapter 13 प्रायिकता विविध प्रश्नावली

Jharkhand Board JAC Class 9 Maths Solutions Chapter 7 Triangles Ex 7.3 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.3

Page-128

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that

(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
Answer:
Given: ΔABC and ΔDBC are two isosceles triangles.
Proof: (i) In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (ΔABC is isosceles)
BD = CD (ΔDBC is isosceles)
Therefore, ΔABD ≅ ΔACD by SSS congruence criterion.

(ii) In ΔABP and ΔACP,
AP = AP (Common)
∠PAB = ∠PAC (ΔABD ≅ ΔACD so by CPCT)
AB = AC (ΔABC is isosceles)
Therefore, ΔABP ≅ ΔACP by SAS congruence criterion.

(iii) ∠PAB = ∠PAC by CPCT as ΔABD ≅ ΔACD.
AP bisects ∠A. ………..(i)
Also, in ΔBPD and ΔCPD,
PD = PD (Common)
BD = CD (ΔDBC is isosceles)
BP = CP (ΔABP ≅ ΔACP so by CPCT)
Therefore, ΔBPD ≅ ΔCPD by SSS congruence criterion.
Thus, ∠BDP = ∠CDP by CPCT. …(ii) By (i) and (ii) we can say that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD (by CPCT as ΔBPD = ΔCPD)
and BP = CP …(hi)
Also, ∠BPD + ∠CPD = 180° (BC is a straight line)
⇒ 2∠BPD = 180°
⇒ ∠BPD = 90° …(iv)
From (i) and (ii),
AP is the perpendicular bisector of BC.

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.

Answer:
Proof: (i) In ΔABD and ΔACD,
∠ADB – ∠ADC = 90°
AB = AC (Given)
AD = AD (Common)
Therefore, ΔABD ≅ ΔACD by RHS congruence criterion.
Now,
BD = CD (by CPCT)
Thus, AD bisects BC.

(ii) ∠BAD = ∠CAD (by CPCT)
Thus, AD bisects ∠A.

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig). Show that:

(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR
Answer:
Given: AB = PQ, BC = QR and AM = PN
Proof:
(i) \(\frac{1}{2}\) BC = BM and \(\frac{1}{2}\) QR = QN (AM and PN are medians)
BC = QR (Given)
⇒ \(\frac{1}{2}\) BC = \(\frac{1}{2}\) QR
⇒ BM = QN

In ΔABM and ΔPQN,
AM = PN (Given)
AB = PQ (Given)
BM = QN (Proved above)
Therefore, ΔABM ≅ ΔPQN by SSS congruence criterion.

(ii) In ΔABC and ΔPQR,
AB = PQ (Given)
∠ABC = ∠PQR
(AABM = APQN so by CPCT) BC = QR (Given)
Therefore, ΔABC ≅ ΔPQR by SAS congruence criterion.

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Ans. Given: BE and CF are two equal altitudes.
Proof: In ABEC and ACFB,
∠BEC = ∠CFB = 90° (Altitudes)
BC = CB (Common)
BE = CF (Given)
Therefore, ΔBEC ≅ ΔCFB by RHS congruence criterion.
Now,
∠C = ∠B (by CPCT)
Thus, AB = AC (sides opposite to the equal angles are equal.)
Therefore, ΔABC is isosceles.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Answer:
Given: AB = AC
Proof: In ΔABP and ΔACP,
AB = AC (Given)
AP = AP (Common)
Therefore, ΔABP ≅ ΔACP by RHS congruence criterion.
Thus, ∠B = ∠C (by CPCT)

Jharkhand Board JAC Class 9 Sanskrit Solutions Shemushi Chapter 2 स्वर्णकाकः Textbook Exercise Questions and Answers.

JAC Board Class 9th Sanskrit Solutions Shemushi Chapter 2 स्वर्णकाकः

JAC Class 9th Sanskrit स्वर्णकाकः Textbook Questions and Answers

1. (अ) एकपदेन उत्तरं लिखत-(एक शब्द में उत्तर लिखिए-)
(क) माता काम् आदिशत्? (माता ने किसको आदेश दिया?)
उत्तरम् :
पुत्रीम् (बेटी को)।

(ख) स्वर्ण-काकः कान् अखादत्? (सोने का कौवा किन्हें खा गया?)
उत्तरम् :
तण्डुलान् (चावलों को)।

(ग) प्रासादः कीदृशः वर्तते? (महल कैसा है?)
उत्तरम् :
स्वर्णमयः (सोने का)।

(घ) गृहमागत्य तया का समुद्घाटिता? (घर आकर उसने क्या खोली?)
उत्तरम् :
मञ्जूषा (पेटी)।

(ङ) लोभाविष्टा बालिका कीदृशीं मञ्जूषां नयति? (लोभी लड़की कैसी मंजूषा को ले जाती है?)
उत्तरम् :
वृहत्तमाम् (सबसे बड़ी को)।

(आ) अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत – (निम्नलिखित प्रश्नों के उत्तर संस्कृत भाषा में लिखिए-)
(क) निर्धनायाः वृद्धायाः दुहिता कीदृशी आसीत् ? (निर्धन वृद्धा की पुत्री कैसी थी?)
उत्तरम् :
निर्धनायाः वृद्धायाः दुहिता विनम्रा मनोहरा चासीत्। (निर्धन वृद्धा की पुत्री विनम्र और मनोहर थी।)

(ख) बालिकया पूर्वं कीदृशः काकः न दृष्टमः आसीत्? (बालिका के द्वारा पहले कैसा कौआ नहीं देखा गया था?)
उत्तरम् :
बालिकया पूर्वं स्वर्णपक्षो रजतचञ्चुः स्वर्णकाकः न दृष्टः आसीत्।
(बालिका के द्वारा पहले सोने के पंखों वाला, चाँदी की चोंच वाला सुनहला कौआ नहीं देखा गया था।)

(ग) निर्धनायाः दुहिता मञ्जूषायां कानि अपश्यत्? (निर्धना की पुत्री ने मंजूषा में किन्हें देखा?)
उत्तरम् :
निर्धनायाः दुहिता मञ्जूषायां महार्हाणि हीरकाणि अपश्यत्। (निर्धन की पुत्री ने मंजूषा में महँगे हीरे देखे।)

(घ) बालिका किं दृष्ट्वा आश्चर्यचकिता जाता? (बालिका क्या देखकर आश्चर्यचकित हो गई?)
उत्तरम् :
बालिका वृक्षस्योपरि स्वर्णमयं प्रासादं दृष्ट्वा आश्चर्यचकिता जाता।
(बालिका वृक्ष के ऊपर सोने का महल देखकर आश्चर्यचकित हो गई।)

(ङ) गर्विता बालिका कीदृशं सोपानम् अयाचत् कीदृशं च प्राप्नोत्?
(घमण्डी बालिका ने कैसी सीढ़ी माँगी. और कैसी प्राप्त की?)
उत्तरम् :
गर्विता बालिका स्वर्णमयसोपानम् अयाचत् ताम्रमयं च प्राप्नोत्।
(घमण्डी बालिका ने सोने की सीढ़ी माँगी और ताँबे की प्राप्त की।)

2. (क) अधोलिखितानां शब्दानां विलोमपदं पाठात् चित्वा लिखत –
(निम्नलिखित शब्दों के विलोमपद पाठ से चुनकर लिखिए-)
(i) पश्चात् …………
(ii) हसितुम् …………
(ii) अधः …………
(iv) श्वेतः …………
(v) सूर्यास्तः …………
(vi) सुप्तः …………
उत्तरम् :
शब्दः – विलोमपदम्
(i) पश्चात् – पूर्वम्
(ii) अधः – उपरि
(iii) सूर्यास्तः – सूर्योदयः
(iv) हसितुम् – रोदितुम्
(v) श्वेतः – कृष्णः
(vi) सुप्तः – प्रबुद्धः

(ख) सन्धिं कुरुत (संधि करिए-)
(i) नि + अवसत् ………
(ii) सूर्य + उदयः ……….
(iii) वृक्षस्य + उपरि ……….
(iv) हि + अकारयत् ……….
(v) च + एकाकिनी ……….
(vi) इति + उक्त्वा ……….
(vii) प्रति + अवदत् ……….
(viii) प्र. + उक्तम् ……….
(ix) अत्र + एव ……….
(x) तत्र + उपस्थिता ……….
(xi) यथा + इच्छम् ……….
उत्तरम् :
(i) न्यवसत्
(ii) सूर्योदयः
(iii) वृक्षस्योपरि
(iv) ह्यकारयत्
(v) चैकाकिनी
(vi) इत्युक्त्वा
(vii) प्रत्यवदत्
(viii) प्रोक्तम्
(ix) अत्रैव
(x) तत्रोपस्थिता
(xi) यथेच्छम्।

3. स्थूलपदान्यधिकृत्य प्रश्ननिर्माणं कुरुत –
(मोटे छपे शब्दों को आधार बनाकर प्रश्न निर्माण कीजिए-)
(क) ग्रामे निर्धना स्त्री अवसत्। (गाँव में निर्धन स्त्री रहती थी।)
(ख) स्वर्णकाकं निवारयन्ती बालिका प्रार्थयत्। (सुनहले कौए को रोकती हुई बालिका ने प्रार्थना की।)
(ग) सूर्योदयात् पूर्वमेव बालिका तत्रोपस्थिता। (सूर्योदय से पूर्व ही बालिका वहाँ उपस्थित हो गई।)
(घ) बालिका निर्धनमातुः दुहिता आसीत्। (बालिका गरीब माता की पुत्री थी।)
(ङ) लुब्धा वृद्धा स्वर्णकाकस्य रहस्यमभिज्ञातवती। (लोभी वृद्धा सुनहले कौए के रहस्य को जान गई।)

प्रश्न:-
(क) ग्रामे का अवसत्? (गाँव में कौन रहती थी?)
(ख) कं निवारयन्ती बालिका प्रार्थयत्? (किसे रोकती हुई बालिका ने प्रार्थना की?)
(ग) कस्मात् पूर्वमेव बालिका तत्रोपस्थिता? (किससे पहले ही बालिका वहाँ उपस्थित हो गई?)
(घ) बालिका कस्याः दुहिता आसीत् ? (बालिका किसकी पुत्री थी?)
(ङ) लुब्धा वृद्धा कस्य रहस्यमभिज्ञातवती? (लोभी वृद्धा किसके रहस्य को जान गई?)

4. प्रकृति-प्रत्यय-संयोगं कुरुत-(प्रकृति-प्रत्यय का संयोग कीजिए)
(पाठात् चित्वा वा लिखत) अथवा पाठ में से छाँटकर लिखिए।
(क) वि + लोक् + ल्यप् – ……………….
(ख) नि + क्षिप् + ल्यप् – ………………..
(ग) आ + गम् + ल्यप् – ………………..
(घ) दृश् + क्त्वा – …………..
(ङ) शी + क्त्वा – ………………
(च) लघु + तमप् – …………..
उत्तर :
(क) विलोक्य
(ख) निक्षिप्य
(ग) आगम्य/आगत्य
(घ) दृष्ट्वा
(ङ) शयित्वा
(च) लघुतमम्

5. प्रकृतिप्रत्यय-विभागं कुरुत-(प्रकृति-प्रत्यय को अलग कीजिए)
(क) रोदितुम् – ……………
(ख) दृष्ट्वा – ……………
(ग) विलोक्य – ……………
(घ) निक्षिप्य – ……………
(ङ) आगत्य – ………………
(च) शयित्वा – ……………
(छ) लघुतमम् – ……………
उत्तर :
(क) रुद् + तुमुन्
(ख) दृश् + क्त्वा
(ग) वि + लोक् + ल्यप्
(घ) नि + क्षिप् + ल्यप्
(ङ) आ + गम् + ल्यप्
(च) शी + क्त्वा
(छ) लघु + तमप्

6. अधोलिखितानि कथनानि कः/का, कं/ काम् च कथयति –
(निम्नलिखित कथनों को कौन किससे कहता है-)

उत्तर :
क:/का – कं/काम्
(क) स्वर्णकाकः (सुनहला कौआ) – बालिकाम् (बालिका से)
(ख) निर्धना वृद्धा (निर्धन वृद्धा) – पुत्रीम् (पुत्री से)
(ग) निर्धना बालिका (निर्धन बालिका) स्वर्णकाकम् (सुनहले कौए से)
(घ) स्वर्णकाकः (सुनहला कौआ) – निर्धनां बालिकाम् (निर्धन बालिका से)
(ङ) गर्विता बालिका (घमण्डी बालिका) – स्वर्णकाकम् (सुनहले कौए से)

7. उदाहरणमनुसृत्य कोष्ठकगतेषु पदेषु पञ्चमीविभक्तेः प्रयोगं कृत्वा रिक्तस्थानानि पूरयत –
(उदाहरणानुसार कोष्ठक में दिये गये पदों में पंचमी विभक्ति का प्रयोग कर रिक्त-स्थानों की पूर्ति कीजिए-) यथा – मूषकः बिलाद् बहिः निर्गच्छति। (बिल)
(क) जनः ……………. बहिः आगच्छति। (ग्राम)
(ख) नद्यः …………….. निस्सरन्ति। (पर्वत)
(ग)……………. पत्राणि पतन्ति। (वृक्ष)
(घ) बालकः ……………. विभेति। (सिंह)
(ङ) ईश्वरः…………………त्रायते। (क्लेश)
(च) प्रभुः भक्तं ………….. निवारयति। (पाप)
उत्तर :
(क) ग्रामात्
(ख) पर्वतेभ्यः
(ग) वृक्षात्
(घ) सिंहात्
(ङ) क्लेशात्
(च) पापात्।

JAC Class 9th Sanskrit स्वर्णकाकः Important Questions and Answers

प्रश्न: 1.
ग्रामे का न्यवसत्? (गाँव में कौन रहती थी?)
उत्तरम् :
ग्रामे एका निर्धना वृद्धा स्त्री न्यवसत्। (गाँव में एक गरीब वृद्धा स्त्री रहती थी।)

प्रश्न: 2.
माता किम् अकरोत्? (माता ने क्या किया?)
उत्तरम् :
माता स्थाल्यां तण्डुलान् निक्षिप्य पुत्री तद्रक्षणाय आदिदेश। (माता ने थाली में चावलों को रखकर पुत्री को उनकी रक्षा करने का आदेश दिया।)

प्रश्न: 3.
किञ्चित् कालादनन्तरं किम् अभवत्? (कुछ समय पश्चात् क्या हुआ?)
उत्तरम् :
किञ्चित् कालादनन्तरम् एकः विचित्रः काकः समुड्डीय तामुपजगाम। (कुछ समय पश्चात् एक विचित्र कौआ उड़कर उसके पास आया।)

प्रश्न: 4.
बालिकया कीदृशः काकः पूर्वं न दृष्टः? (बालिका ने कैसा कौआ पहले नहीं देखा था?)
उत्तरम् :
बालिकया स्वर्णपक्षो रजतचञ्चुः स्वर्णकाकः पूर्वं न दृष्टः। (बालिका ने सोने के पंखों वाला चाँदी की चोंच वाला सुनहला कौआ पहले नहीं देखा था।) .

प्रश्न: 5.
बालिका काकं किं प्रार्थयत्? (बालिका ने कौए से क्या प्रार्थना की?)
उत्तरम् :
बालिका काकं प्रार्थयत्-तण्डुलान् मा भक्षय। मदीया माता अतीव निर्धना वर्तते। (बालिका ने कौए से प्रार्थना की-चावलों को मत खाओ। मेरी माता बहुत गरीब है।)

प्रश्न: 6.
किं विलोक्य बालिका रोदितुम् आरब्धा? (क्या देखकर ..लिका ने रोना प्रारम्भ कर दिया?)
उत्तरम् :
काकं तण्डुलान् खादन्तं हसन्तञ्च विलोक्य बालिका रोदितुम् आरब्धा। . (कौए को चावलों को खाते हुए और हँसते हुए देखकर बालिका ने रोना आरम्भ कर दिया।)

प्रश्न: 7.
बालिका (कन्या) ग्रामात बहिः कदा उपस्थिता? (बालिका गाँव से बाहर कब उपस्थित हो गई?)
उत्तरम् :
बालिका सूर्योदयात्पूर्वमेव ग्रामात् बहिः उपस्थिता। (बालिका सूर्योदय से पूर्व ही गाँव से बाहर उपस्थित हो गई।)

प्रश्न: 8.
बालिकां वृक्षस्य अधः दृष्ट्वा काकेन किं कथितम्? (वृक्ष के नीचे बालिका को देखकर कौए ने क्या कहा?)
उत्तरम् :
काकेन कथितम्-हं हो. बाले! त्वमागता, तिष्ठ, अहं त्वत्कृते सोपानमवतारयामि, तत्कथय स्वर्णमयं रजतमयमुत ताम्रमयं वा। (कौए ने कहा-अरे बालिके तुम आ गई, ठहरो, मैं तुम्हारे लिए सीढ़ी उतारता हूँ, तो कहो सोने की या चाँदी की अथवा ताँबे की।)

प्रश्न: 9.
सोपानविषये कन्या काकं किं प्रावोचत? (सीढी के विषय में बालिका ने कौए से क्या कहा?)
उत्तरम् :
कन्या काकं प्रावोचत्-अहं निर्धनमातुर्दुहिता अस्मि। ताम्रसोपानेनैव आगमिष्यामि। (बालिका ने कौए से कहा-मैं निर्धन माता की पुत्री हूँ। ताँबे की सीढ़ी से ही आऊँगी।)

प्रश्न: 10.
श्रान्तां कन्यां विलोक्य काकः किं प्राह? (कन्या को थकी हुई देखकर कौए नें क्या कहा?)
उत्तरम् :
कन्यां श्रान्तां विलोक्य काकः प्राह-पूर्वं लघुप्रातराशः क्रियताम्। (बालिका को थकी हुई देखकर कौआ बोला-पहले थोड़ा-सा सुबह का नाश्ता कर लो।)

प्रश्न: 11.
बालिका स्थाल्याविषये किं व्याजहार? (बालिका ने थाली के विषय में क्या कहा?)
उत्तरम् :
स्थाल्याविषये बालिका व्याजहार-ताम्रस्थाल्यामेवाहं निर्धना भोजनं करिष्यामि। (थाली के विषय में बालिका ने कहा-मैं गरीब हूँ, (इसलिए) ताँबे की थाली में ही भोजन करूँगी।)

प्रश्न: 12.
प्रातराशकाले कन्या कदा आश्चर्यचकिता जाता? (नाश्ते के समय बालिका कब आश्चर्यचकित हो गई?)
उत्तरम् :
कन्या तदा आश्चर्यचकिता सञ्जाता यदा स्वर्णकाकेन स्वर्णस्थाल्यां भोजनं परिवेषितम्। (बालिका तब आश्चर्यचकित हो गई जब सुनहले कौए ने सोने की थाली में भोजन परोसा।)

प्रश्न: 13.
अन्ते काकः किं ब्रूते? (अन्त में कौए ने क्या कहा?)
उत्तरम् :
अन्ते काक: ब्रूते-बालिके! अहम् इच्छामि यत् त्वं सर्वदा अत्रैव तिष्ठ परं तव माता एकाकिनी वर्तते। (अन्त में कौए ने कहा-बालिका! मैं चाहता हूँ कि तुम हमेशा यहीं पर ठहरो लेकिन तुम्हारी माँ अकेली है।)

प्रश्न: 14.
काकः कक्षाभ्यन्तरात् किम् आनयत्? (कौआ कमरे के अन्दर से क्या लाया?)
उत्तरम् :
काकः कक्षाभ्यन्तरात् तिस्रः मञ्जूषाः आनयत्। (कौआ कमरे के अन्दर से तीन सन्दूकें लाया।)

प्रश्न: 15.
मञ्जूषाः निस्सार्य काकः बालिकां किम् अवदत्? (सन्दूकें निकालकर कौए ने बालिका से क्या कहा?)
उत्तरम् :
काकः बालिकाम् अवदत्-बालिके! यथेच्छं गृहाण मञ्जूषामेकाम्। (कौआ बालिका से बोला-बालिका! इच्छानुसार एक सन्दूक ग्रहण करो।)

प्रश्न: 16.
बालिका का मञ्जूषां गृहीतवती? (बालिका ने कौनसा सन्दूक लिया?)
उत्तरम् :
बालिका लघुतमां मञ्जूषां गृहीतवती। (बालिका ने सबसे छोटा सन्दूक लिया।)

प्रश्न: 17.
मञ्जूषां प्रगृह्य बालिकया किं कथितम्? (सन्दूक को लेकर बालिका ने क्या कहा?)
उत्तरम् :
मञ्जूषा प्रगृह्य बालिकया कथितम्-इयत् एव मदीयतण्डुलाना मूल्यम्। (सन्दूक लेकर बालिका ने कहा-इतना ही मेरे चावलों का मूल्य है।)

प्रश्न: 18.
मञ्जूषां समुद्घाट्य बालिका तस्यां किम् अपश्यत्? (सन्दूक को खोलकर बालिका ने उसमें क्या देखा?)
उत्तरम् :
मञ्जूषां समुद्घाट्य बालिका तस्यां महार्हाणि हीरकाणि अपश्यत्। (सन्दूक को खोलकर बालिका ने उसमें बहुमूल्य हीरों को देखा।)

प्रश्न: 19.
अपरा वृद्धा कस्य रहस्यम् अभिज्ञातवती? (दूसरी वृद्धा किसके रहस्य को जान गई?)
उत्तरम् :
अपरा वृद्धा स्वर्णकाकस्य रहस्यम् अभिज्ञातवती। (दूसरी वृद्धा सुनहले कौए के रहस्य को जान गई।)

प्रश्न: 20.
काकः निर्धनायै बालिकायै किं दातुम् अकथयत्? (कौए ने निर्धन बालिका के लिए क्या देने को कहा?)
उत्तरम् :
काकः निर्धनायै बालिकायै तण्डुलमूल्यं दातुम् अकथयत्। (कौए ने निर्धन बालिका के लिए चावलों का मूल्य देने को कहा।)

प्रश्न: 21.
निर्धना कन्या कीदृशं सोपानं अवतारयितुम् अकथयत्? (गरीब कन्या ने कैसी सीढ़ी उतारने को कहा?)
उत्तरम् :
निर्धना कन्या ताम्रसोपानम् अवतारयितुम् अकथयत्। (गरीब कन्या ने ताँबे की सीढ़ी उतारने को कहा।)

प्रश्न: 22.
निर्धनां बालिका काकः कति मञ्जूषाः ग्रहीतुम् अकथयत्? (कौए ने निर्धन बालिका से कितनी सन्दूकें लेने को कहा?)
उत्तरम् :
काकः निर्धनां बालिकाम् एकां मञ्जूषां ग्रहीतुम् अकथयत्। (कौए ने निर्धन बालिका से एक सन्दूक लेने को कहा।) ।

प्रश्न: 23.
लुब्धया बालिकया का मञ्जूषा गृहीता? (लोभी बालिका ने कौन-सी सन्दूक ग्रहण की?)
उत्तरम् :
लुब्धया बालिकया बृहत्तमा मञ्जूषा गृहीता। (लोभी बालिका ने सबसे बड़ी सन्दूक ग्रहण की।)

रेखांकित पदान्यधिकृत्य प्रश्न निर्माणं कुरुत-(रेखांकित पदों के आधार पर प्रश्न निर्माण करिए-)

प्रश्न: 1.
स्वर्णकाकः इति कथायां लोभस्य परिणामो वर्णितः। (स्वर्णकाकः कहानी में लोभ का परिणाम वर्णित है।)
उत्तरम् :
स्वर्णकाकः इति कथायां कस्य परिणामो वर्णित:? (इस कहानी में किसका परिणाम वर्णित है?)

प्रश्न: 2.
दुहिता विनम्रा मनोहरा चासीत्। (बेटी विनम्र और मनोहर थी।)
उत्तरम् :
दुहिता कीदृशी आसीत् ? (बेटी कैसी थी?)

प्रश्न: 3.
स्थाल्यां तण्डुलान् निक्षिप्य पुत्रीम् आदिशत्। (थाली में चावल रखकर बेटी को आदेश दिया।)
उत्तरम् :
स्थाल्यां कान् निक्षिप्य पुत्रीम् आदिशत्? (थाली में किन्हें रखकर बेटी को आदेश दिया।)

प्रश्न: 4.
बालिका रोदितुमारब्धा। (बालिका ने रोना आरम्भ किया।)
उत्तरम् :
का रोदितुमारब्धा? (किसने रोना आरम्भ किया?)

प्रश्नः 5.
सूर्योदयात्पूर्वमेव सा तत्रोपस्थिता। (सूर्योदय से पूर्व ही वह वहाँ उपस्थित हो गई।)
उत्तरम् :
सा तत्र कदा उपस्थिता? (वह वहाँ कब उपस्थित हुई?)

प्रश्न: 6.
सा स्वर्णसोपानेन स्वर्णभवनम् आरोहत। (वह सोने की सीढ़ी से स्वर्ण-भवन पर चढ़ी।)
उत्तरम् :
सा केन स्वर्णभवनम् आरोहत? (वह किससे सोने के भवन पर चढ़ी?)

प्रश्नः 7.
तदा सा आश्चर्यचकिता सजाता? (तब वह आश्चर्यचकित हो गई।)
उत्तरम् :
तदा सा कीदृशी सञ्जाता? (तब वह कैसी हो गई?)

प्रश्नः 8.
स्वर्णकाकेन स्वर्णस्थाल्यां भोजनं परिवेषितम्। (स्वर्ण काक द्वारा सोने की थाली में भोजन परोसा गया।)
उत्तरम् :
स्वर्णकाकेन कस्यां भोजनं परिवेषितम्? (स्वर्ण काक ने किसमें भोजन परोसा?)

प्रश्न: 9.
काकः कक्षाभ्यन्तरात् तिस्त्रः मञ्जूषा निस्सार्य अददात्। (कौए ने कमरे के अन्दर से तीन पेटी निकालकर दी।)
उत्तरम् :
काकः कुतः तिस्त्रः मञ्जूषा निस्सार्य अददात्। (कौए ने कहाँ से तीन पेटी निकालकर दी।)

प्रश्न: 10.
लुब्धया बालिकया लोभस्य फलं प्राप्तम्? (लोभी बालिका द्वारा लोभ का फल प्राप्त किया गया।)
उत्तरम् :
कया लोभस्य फलं प्राप्तम्? (किसके द्वारा लोभ का फल प्राप्त किया गया?)
कथाक्रम-संयोजनम् अधोलिखितवाक्यानि पठित्वा कथाक्रमसंयोजनं कुरुत (निम्नलिखित वाक्यों को पढ़कर कथा-क्रम-संयोजन कीजिए)

1. ईय॑या सा तस्य स्वर्णकाकस्य रहस्यमभिज्ञातवती।
2. तस्मिन्नेव ग्रामे एकाऽपरा लुब्धा वृद्धा न्यवसत्।
3. लघुतमा मञ्जूषां प्रगृह्य बालिकया कथितमियदेव मदीयतण्डुलानां मूल्यम्।
4. काकः प्राह-पूर्वं लघुप्रातराशः क्रियताम्।
5. भवने चित्रविचित्रवस्तूनि सज्जितानि दृष्ट्वा सा विस्मयं गता।
6. सूर्योदयात्पूर्वमेव सा तत्रोपस्थिता।
7. नैतादृशः स्वर्णपक्षो रजतचञ्चुः स्वर्णकाकस्तया पूर्वं दृष्टः।
8. तस्याश्चैका दुहिता विनम्रा मनोहरा चासीत्।

उत्तरम् :

1. तस्याश्चैका दुहिता विनम्रा मनोहरा चासीत्।
2. नैतादृशः स्वर्णपक्षो रजतचञ्चुः स्वर्णकाकस्तया पूर्वं दृष्टः।
3. सूर्योदयात्पूर्वमेव सा तत्रोपस्थिता।
4. भवने चित्रविचित्रवस्तूनि सज्जितानि दृष्ट्वा सा विस्मयं गता।
5. काकः प्राह-पूर्वं लघुप्रातराशः क्रियताम्।
6. लघुतमा मञ्जूषां प्रगृह्य बालिकया कथितमियदेव मदीयतण्डुलानां मूल्यम्।
7. तस्मिन्नेव ग्रामे एकाऽपरा लुब्धा वृद्धा न्यवसत्।
8. ईय॑या सा तस्य स्वर्णकाकस्य रहस्यमभिज्ञातवती।

भाषिक-विस्तार –

1. किसी भी काम को करके’ इस अर्थ में ‘क्त्वा’ प्रत्यय का प्रयोग किया जाता है। यथा-
पठित्वा – पठ् + क्त्वा = पढ़कर
गत्वा – गम् + क्त्वा = जाकर
खादित्वा – खाद् + क्त्वा = खाकर

इसी अर्थ में अगर धातु (क्रिया) से पहले उपसर्ग होता है तो ल्यप् प्रत्यय का प्रयोग होता है। धातु से पूर्व उपसर्ग होने की स्थिति में कभी भी ‘क्त्वा’ प्रत्यय नहीं हो सकता और उपसर्ग न होने की स्थिति में कभी भी ल्यप् प्रत्यय नहीं हो सकता है। यथा –

उप + गम् + ल्यप् = उपगम्य
सम् + पूज् + ल्यप् = सम्पूज्य
वि + विलोक्य आ + ल्यप् = आदाय
निर् + गम् + ल्यप् = निर्गत्य

2. प्रश्नवाचक शब्दों को अनिश्चयवाचक बनाने के लिए चित् और चन निपातों का प्रयोग किया जाता है। ये निपात जब सर्वनाम पदों के साथ लगते हैं तो सर्वनाम पद होते हैं और जब अव्यय पदों के साथ प्रयुक्त होते हैं तो अव्यय होते हैं।

यथा – कः = कौन
कः + चन = कश्चन = कोई
कः + चित् = कश्चित् = कोई
के + चन = केचन = कोई (बहुवचन में)
के + चित् = केचित् (बहुवचन में)
का + चन = काचन = कोई (स्त्री)
का + चित् = काचित् (कोई स्त्री)
काः + चन = काश्चन = कुछ स्त्रियाँ (बहुवचन में)
काः+ चित् = काश्चित् (कुछ स्त्रियाँ) (बहुवचन में)

“किम्’ शब्द के सभी वचनों, लिंगों व सभी विभक्तियों में ‘चित्’ और ‘चन’ का प्रयोग किया जा सकता है और उसे अनिश्चयवाचक बनाया जा सकता है।

जैसे- (i) किञ्चित्-प्रथमा में
(ii) केनचित्-तृतीया में
(iii) केषाञ्चित् (केषाम् + चित्)-षष्ठी में
(iv) कस्मिंश्चित्-सप्तमी में
(v) कस्याञ्चित्-सप्तमी (स्त्रीलिङ्ग में)

इस तरह ‘चित्’ के स्थान पर ‘चन’ का प्रयोग होता है। ‘चित्’ और ‘चन’ जब अव्यय पदों में लग जाते हैं तो वे अव्यय हो जाते हैं। जैसे- क्वचित्-क्वचन कदाचित्-कदाचन

3. संस्कृत में एक से चतुर् (चार) तक संख्यावाची शब्द पुल्लिङ्ग, स्त्रीलिङ्ग तथा नपुंसकलिङ्ग में अलग-अलग रूपों में होते हैं पर पञ्च (पाँच) से उनका रूप सभी लिङ्गों में एक-सा होता है।

स्वर्णकाकः Summary and Translation in Hindi

पाठ परिचयः – प्रस्तुत पाठ श्री पद्मशास्त्री द्वारा विरचित ‘विश्वकथाशतकम्’ नामक कथासंग्रह से लिया गया है। जिसमें विभिन्न देशों की सौ लोक कथाओं का संग्रह है। यह वर्मा देश की श्रेष्ठतम कथाओं में से एक है, जिसमें लोभ और उसके दुष्परिणामों के साथ-साथ त्याग और उसके सुपरिणामों का वर्णन एक ‘सुनहले पंखों वाले कौए’ के माध् यम से किया गया है।

प्राचीन समय में एक गाँव में एक वृद्ध स्त्री रहती थी। उसकी पुत्री विनम्र और सुन्दर थी। एक बार माता ने उससे धूप में रखे चावलों की पक्षियों से रक्षा करने को कहा। कुछ समय पश्चात् एक सुनहले पंखों वाला कौआ उड़ता हुआ वहाँ पहुँचा और चावलों को खाने लगा।

लड़की के मना करने पर उस कौए ने उस लड़की से सूर्योदय से पहले गाँव के बाहर पीपल के पेड़ के नीचे आकर चावलों का मूल्य लेने को कहा। जब सूर्योदय से पहले वह लड़की वहाँ पहुँची तो उस लड़की ने वृक्ष के ऊपर सोने का महल देखा। लड़की की त्याग-भावना से प्रसन्न हुए कौए ने उस लड़की को सोने की सीढ़ी से ऊपर चढ़ाया तथा सोने की थाली में स्वादिष्ट भोजन कराया। लड़की ने कौए द्वारा दी गयी तीन सन्दूकों में से सबसे छोटी सन्दूक को चावलों के मूल्य के रूप में स्वीकार किया। लड़की ने घर पहुँचकर सन्दूक को खोलकर देखा, उस सन्दूक में भरे हीरों को देखकर वह बहुत प्रसन्न हुई और वह उसी दिन से धनिक बन गई।

उसी गाँव की एक लालची स्त्री को जब उस निर्धन लड़की के धनिक बनने का पता चला तो उस लालची स्त्री ने भी अपनी पुत्री से धूप में रखे चावलों की पक्षियों से रक्षा करने को कहा। जब वह सुनहला कौआ वहाँ पहुँचकर चावलों को खाने लगा तो उस कौए को लड़की ने भी उन चावलों को खाने से रोका। कौए ने उस लड़की से भी सूर्योदय से पूर्व ! गाँव के बाहर पीपल के वृक्ष के नीचे आने को कहा। वह लड़की सूर्योदय से पहले वहाँ पहुँचकर कौए की निन्दा करने । लगी और चावलों का मूल्य माँगने लगी।

कौए द्वारा सीढ़ियों की पूछने पर अभिमानी लड़की बोली कि सोने की सीढ़ियों से आऊँगी, परन्तु ‘सुनहले पंखों वाले कौए’ ने उसके लिए ताँबे की सीढ़ी दी। कौए ने उस लड़की को भोजन भी ताम्रपात्र में कराया। कौए द्वारा दी गई तीन पेटियों में से सबसे बड़ी पेटी लेकर वह खुशी-खुशी घर लौटी और जब उस लड़की ने उस पेटी को खोलकर देखा तो भयंकर काले साँप के रूप में अपने लोभ के फल को प्राप्त कर दुःखी हुई। इसके पश्चात् उसने लालच करना छोड़ दिया।

मूलपाठः, शब्दार्थाः, सप्रसंग हिन्दी-अनुवादः, सप्रसंग संस्कृत-व्यारव्याः अवबोधन कार्यम् च । __ 1. पुरा कस्मिंश्चिद् ग्रामे एका निर्धना वृद्धा स्त्री न्यवसत्। तस्याः च एका दुहिता विनम्रा मनोहरा चासीत्। एकदा माता स्थाल्यां तण्डुलान् निक्षिप्य पुत्रीम् आदिशत्। “सूर्यातपे तण्डुलान् खगेभ्यो रक्षा” किञ्चित् कालादनन्तरम् एको विचित्रः काकः समुड्डीय तस्याः समीपम् अगच्छत्।

नैतादृशः स्वर्णपक्षो रजतचञ्चुः स्वर्णकाकस्तया पूर्वं दृष्टः। तं तण्डुलान् खादन्तं हसन्तञ्च विलोक्य बालिका रोदितुमारब्धा। तं निवारयन्ती सा प्रार्थयत्- “तण्डुलान् मा भक्षया मदीया माता अतीव निर्धना वर्तते।” स्वर्णपक्षः काकः प्रोवाच, “मा शुचः। सूर्योदयात्प्राग् ग्रामाद्बहिः पिप्पलवृक्षमनु त्वया आगन्तव्यम्। अहं तुभ्यं तण्डुलमूल्यं दास्यामि।” प्रहर्षिता बालिका निद्रामपि न लेभे।

शब्दार्थाः – पुरा = प्राचीनकाले (प्राचीन समय में), कस्मिंश्चिद् = कस्मिन् चित् (किसी), ग्रामे = वसथे (गाँव में), एका निर्धना = एका दरिद्रा (एक गरीब), स्त्री = वनिता (महिला/स्त्री), न्यवसत् = वसति स्म (रहती थी), तस्याः = (उसकी), एका दुहिता = एका पुत्री (एक पुत्री), विनम्रा = विनयशीला (विनम्र), मनोहरा = हृदयहारिणी (मनोहर), च = और, आसीत् = अवर्तत (थी), एकदा = एकस्मिन् दिवसे (एक बार), माता = जननी (माता ने), स्थाल्याम् = स्थालीपात्रे (थाली में), तण्डुलान् = अक्षतान् (चावलों को), निक्षिप्य = प्रसार्य (फैलाकर/रखकर), पुत्रीम आदिशत् = आदेशमददत् (आदेश दिया), सूर्यातपे = आतपे (धूप में),

तण्डुलान् = अक्षतान् (चावलों को), खगेभ्यः = विहगेभ्यः (पक्षियों से), रक्ष = रक्षणं कुरु (रक्षा करो), किञ्चित् कालादनन्तरम् = ईषत् समयपश्चात् (कुछ समय के पश्चात्), एको विचित्रः काकः = एकः विचित्र: वायसः (एक अजीब कौआ), समुड्डीय = उत्प्लुत्य (उड़कर), तस्याः समीपम् अगच्छत् = ताम् समीपम् आगतवान् (उसके पास पहुँचा), न = मा (नहीं), एतादृशः = एतस्य प्रकारस्य (इस प्रकार का), स्वर्णपक्षः = स्वर्णमये पक्षे यस्य सः (सोने के पंखों वाला), रजतचञ्चुः = रजतमय: चञ्चुः यस्य सः (चाँदी की चोंच वाला), स्वर्णकाकः = स्वर्णस्य वायसः (सोने का कौआ/सुनहला कौआ), तया = (उस (लड़की) के द्वारा),

पूर्वम् = पुरा (पहले), दृष्टः = अपश्यत् (देखा था), तम् = उस (कौए) को), तण्डुलान् = अक्षतान् (चावलों को), खादन्तम् = भक्षयन्तम् (खाते हुए), हसन्तम् = हास्यं कुर्वन्तम् (हँसते हुए), च = और, विलोक्य = दृष्ट्वा (देखकर), बालिका = कन्या (उस लड़की ने), रोदितुम् = विलपितुम् (रोना), आरब्धा = प्रारम्भम् अकरोत् (शुरू कर दिया), तम् = (उस कौए को), निवारयन्ती = वारणं कुर्वन्ती (रोकती हुई), सा = (वह), प्रार्थयत् = न्यवेदयत् (प्रार्थना करने लगी), तण्डुलान् = अक्षतान् (चावलों को), मा = न (नहीं/मत), भक्षय = खाद (खाओ), मदीया = मम (मेरी), माता = जननी (माता), अतीव = अत्यन्तम् (अत्यन्त/अत्यधिक),

निधन = वित्तहीना/दरिद्रा (गरीब), वर्तते = अस्ति (है), स्वर्णपक्षः काकः = स्वर्णमये पक्षे यस्य सः (सोने के पंखों वाला/सुनहला कौआ), प्रोवाच = अकथयत् (कहा), मा शुचः = शोक मा कुरु (शोक मत करो), सूर्योदयात्प्राग् = भानूदयात् पूर्वम् (सूर्योदय से पहले), ग्रामाबहिः = गाँव के बाहर, पिप्पलवृक्षमनु = पिप्पलतरो: अधः (पीपल के पेड़ के नीचे), त्वया = भवत्या (तुम्हारे द्वारा), आगन्तव्यम् = आगच्छेत् (आना चाहिए), अहम् = मैं, तुभ्यम् = ते/भवत्यै (तुम्हारे लिए), तण्डुलमूल्यम् = अक्षतमूल्यम् (चावलों का मूल्य), दास्यामि = अर्पणं करिष्यामि (दूँगा), प्रहर्षिता = प्रसन्ना (खुश हुई), बालिका = कन्या (लड़की ने), निद्रामपि न लेभे = शयनम् अपि न प्राप्तम् अकरोत् (नींद भी नहीं ली/सोई भी नहीं)।

हिन्दी-अनुवादः

सन्दर्भ – प्रस्तुत गद्यांश श्रीपद्मशास्त्री द्वारा रचित ‘विश्वकथाशतकम्’ नामक कथा-संग्रह से उद्धृत ‘स्वर्णकाक:’. नामक पाठ से अवतरित है।

प्रसंग – प्रस्तुत गद्यांश में धूप में रखे बालिका द्वारा रक्षित चावलों को सुनहले कौए द्वारा खाने और चावलों का मूल्य चुकाने के लिए कहने का वर्णन है।

हिन्दी-अनुवाद – प्राचीन समय में किसी गाँव में एक गरीब वृद्ध महिला रहती थी। उसकी एक विनम्र और मनोहर पुत्री थी। एक बार माता ने थाली में चावलों को रखकर पुत्री को आदेश दिया-धूप में रखे चावलों की पक्षियों से रक्षा करना। कुछ समय पश्चात् एक अजीब कौआ उड़कर उसके पास पहुँचा।

इस प्रकार का सोने के पंखों वाला, चाँदी की चोंच वाला सुनहला कौआ उसने पहले नहीं देखा था। उसको चावलों को खाता हुआ और हँसता हुआ देखकर लड़की ने रोना शुरू कर दिया। उसको रोकते हुए वह प्रार्थना करने लगी-चावलों को मत खाओ। मेरी माता बहुत गरीब है। सुनहला कौआ बोला-दुखी मत होओ। तुम सूर्योदय से पहले गाँव के बाहर पीपल के पेड़ के नीचे आना। मैं तुम्हें चावलों का मूल्य दूंगा। खुश (प्रसन्न) हुई लड़की को नींद भी नहीं आई अर्थात् वह सोई भी नहीं।

संस्कत-व्याख्याः

सन्दर्भ: – प्रस्तुतगद्यांशः अस्माकं पाठ्यपुस्तकात् शेमुष्याः ‘स्वर्णकाकः’ नामक पाठात् अवतरितोऽस्ति। प्रस्तुतपाठः श्री पद्मशास्त्रिणा विरचितात् ‘विश्वकथाशतकम्’ नामक कथासंग्रहात् सङ्कलितः अस्ति। (प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक. ‘शेमुषी’ के ‘स्वर्णकाकः’ नामक पाठ से लिया गया है। प्रस्तुत पाठ श्री पद्म शास्त्री द्वारा रचित ‘विश्वकथाशतकम्’ नाम के कथा ग्रन्थ से संकलित है।)

प्रसङ्गः – प्रस्तुतगद्यांशे वर्णितं यदेक: स्वर्णकाकः सूर्यातपे स्थाल्यां स्थितान् बालिकया रक्षितान् अक्षतान् खादति तेषां मूल्यं च प्रदातुं कथयति। (प्रस्तुत गद्यांश में वर्णित है कि एक सोने का कौआ धूप में थाली में रखे चावलों को खाता है और उनका मूल्य चुकाने के लिए कहता है।)

व्याख्याः – प्राचीनकाले कस्मिन् अपि वसथे (ग्रामे) एका दरिद्रा वृद्धा वनिता (स्त्री) वसति स्म। तस्याः एका विनयशीला हृदयहारिणी च पुत्री आसीत्। एकस्मिन् दिवसे जननी स्थाल्यां अक्षतान् प्रसार्य सुतायै आदेशम् अददात् यत् सूर्यातपे अक्षतानां विहगेभ्यः रक्षणं कुरु। ईषत् समयपश्चात् एकः विचित्रवायसः उत्प्लुत्य तस्याः समीपम् आगच्छत् । न तस्य प्रकारस्य हेममयः पक्षः रजतमयः चञ्चुः स्वर्णकाकः एतया पुरा दृष्टः। तम् (स्वर्णकाकम्) अक्षतान् भक्षयन्तम् हास्यं च कुर्वन्तं दृष्ट्वा सा कन्या विलपितुम् आरभत।

काकं वारयन्ती सा न्यवेदयत्- अक्षतान् मा खाद। मम जननी अत्यन्त दरिद्रा अस्ति। हेममयः पक्षः वायसः अकथयत्-शोकं मा कुरु। भानूदयात् पूर्वं ग्रामात् बहिः पिप्पलतरा: अधः भवती आगच्छे। अहं भवत्यै अक्षतमूल्यम् अर्पणं करिष्यामि। प्रसन्ना कन्या शयनम् अपि न प्राप्तम् अकरोत्। (प्राचीनकाल में किरा में एक गरीब बुढ़िया रहती थी। उसकी एक विनम्र, हृदय को हरण करने वाली बेटी थी। एक दिन माता ने थाली में चावलों को फैलाकर पुत्री को आदेश दिया कि सूर्य की धूप तक चावलों की पक्षियों से रक्षा करो।

कुछ समय बाद एक विचित्र कौआ उड़कर उसके पास आया। इस प्रकार का सोने के पंखों वाला और चाँदी की चोंच वाला कौआ इससे पहले नहीं देखा गया। उस (सोने के कौवे) को चावल चुगते हुए और हँसते हुए को देखकर वह कन्या रोने लगी। कौए को रोकती हुई उसने निवेदन किया-“चावलों को मत खाओ। मेरी माता बहुत गरीब है।” सोने की चोंच वाले कौवे ने कहा-“दुखी मत हो। सूरज निकलने से पहले गाँव के बाहर पीपल के नीचे आप आ जाना। मैं आपके लिए चावलों का मूल्य सौंप दूंगा।” खुश हुई कन्या सो भी नहीं पाई।)

अवबोधन कार्यम्

प्रश्न 1.
एकपदेन उत्तरत- (एक शब्द में उत्तर दीजिए-)
(क) बालिका तण्डुलान् केभ्यः रक्षेतु? (बालिका किनसे चावलों की रक्षा करे?)
(ख) ‘तण्डुलान् मा खादय’ इति कोऽवदत्? (‘चावल मत खाओ’ ऐसा किसने कहा?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत – (पूरे वाक्य में उत्तर दीजिए-)
(क) बालिका केन कारणेन रोदितुम् आरब्धा? (बालिका किस कारण से रोई ?)
(ख) कीदृशः काकः बालिकया पूर्वं न दृष्टः? (बालिका ने कैसा कौआ पहले नहीं देखा?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) “स्वर्णपक्षी’ इत्यत्र किं पदं विशेषण पदम् ?। (‘स्वर्णपक्षी’ इनमें कौनसा पद विशेषण पद है?)
(ख) ‘धनविहीना’ इति पदस्य समानार्थी गद्यांशात् चित्वा लिखत।
(‘धनविहीना’ इस पद का समानार्थी पद गद्यांश से लिखिए।)
उत्तराणि :
(1) (क) खगेभ्यः। (पक्षियों से)।
(ख) बालिका (लड़की ने)।

(2) (क) तण्डुलान् खादन्तं हसन्तं तं काकम् अवलोक्य बालिका रोदितुम् आरब्धा। (चावलों को खाते हुए हँसते हुए उस कौआ को देखकर लड़की रोने लगी।)
(ख) नैतादृशः स्वर्णपक्षो रजतचञ्चुः स्वर्ण काकस्तया पूर्व दृष्टः। (ऐसा सोने की पंखों और चाँदी की चोंच वाला सोने का कौए उसने पहले नहीं देखा था।)

(3) (क) स्वर्ण (सोना)।
(ख) निर्धना (गरीब)।

2. सूर्योदयात्पूर्वमेव सा तत्रोपस्थिता। वृक्षस्योपरि विलोक्य सा च आश्चर्यचकिता सञ्जाता यत् तत्र स्वर्णमयः प्रासादो वर्तते। यदा काकः शयित्वा प्रबुद्धस्तदा तेन स्वर्णगवाक्षात्कथितं “हहो बाले! त्वमागता, तिष्ठ, अहं त्वत्कृते सोपानमवतारयामि, तत्कथय स्वर्णमयं रजतमयम् ताम्रमयं वा”? कन्या अवदत् “अहं निर्धनमातुः दुहिता अस्मि। ताम्रसोपानेनैव आगमिष्यामि।” परं स्वर्णसोपानेन सा स्वर्ण-भवनम् आरोहता।

शब्दार्थाः – सूर्योदयात्यूर्वमेव = भानूदयात् प्राग् एव (सूर्योदय से पहले ही), सा = (वह (बालिका)), तत्रोपस्थिता = तस्मिन् स्थाने सन्निहिता (वहाँ उपस्थित हो गयी), वृक्षस्योपरि = तरोः उपरि (पेड़ के ऊपर), विलोक्य = दृष्ट्वा (देखकर), सा = (वह), च = (और), आश्चर्यचकिता = विस्मिता (आश्चर्यचकित), सञ्जाता = अभवत् (हो गयी), यत् = (कि), तत्र = तस्मिन् स्थाने (उस जगह/वहाँ) स्वर्णमयः = हेममयः (सोने का), प्रासादः = राजमन्दिरम् (महल), वर्तते = अस्ति (है), यदा = यस्मिन् समये (जब),

काकः = वायसः (कौआ), शयित्वा = शयनं कृत्वा (सोकर), प्रबुद्धः = जागरणम् अकरोत् (जागा), तदा = तस्मिन् समये (तब), तेन = काकेन (उस कौए ने), स्वर्णगवाक्षात् = हेममयवातायनात् (सोने की खिड़की से), कथितं = अकथयत् (कहा), हं हो बाले! = भो बाले (अरे बालिका), त्वमागता = भवती आगता (तुम आ गई हो), तिष्ठ = स्थिरा भव (ठहरो), अहम् = मैं, त्वत्कृते = तुभ्यम् (तुम्हारे लिए), सोपानम् = (सीढ़ी), अवतारयामि = अवतीर्णं करोमि (उतारता हूँ), तत्कथय = तु वद (तो बोलो),

स्वर्णमयं = हेममयं (सोने से बनी, सोने की), रजतमयं = रूप्यमयं (चाँदी से बनी, चाँदी की), वा = (अथवा), ताम्रमयं = लोहितायसमयं (ताँबे से बनी, ताँबे की), कन्या = बालिका (लड़की), अवदत् = प्रावोचत् (बोली), अहं निर्धनमातुः = अहं दरिद्रायाः जनन्याः (मैं निर्धन/गरीब माँ, ताम्रसोपानेनैव = लोहितायसेन निर्मितेन सोपानेन एव (ताँबे की बनी सीढ़ी से ही), आगमिष्यामि = आगमनं करिष्यामि (आऊँगी), परं = परञ्च (लेकिन), स्वर्णसोपानेन = हेमनिर्मितसोपानेन (सोने की सीढ़ी से), स्वर्णभवनम् = हेममयं भवनम् (सोने के महल को), सा = (वह), आरोहत = आरूढवती

हिन्दी-अनुवादः

सन्दर्भ – प्रस्तुत गद्यांश श्री पद्मशास्त्री द्वारा रचित ‘विश्वकथाशतकम्’ नामक कथा-संग्रह से उद्धृत ‘स्वर्णकाकः’ नामक पाठ से अवतरित है।

प्रसंग – प्रस्तुत गद्यांश में बालिका के सूर्योदय से पहले ही गाँव के बाहर पीपल के पेड़ के नीचे पहुँचकर पेड़ के ऊपर सोने के महल को देखकर आश्चर्यचकित होने एवं सोने की सीढ़ियों से स्वर्णमहल में पहुँचने का वर्णन है।

हिन्दी-अनुवाद – सूर्योदय से पहले ही वह (लड़की) वहाँ आ गयी। पेड़ के ऊपर देखकर वह आश्चर्यचकित हो गयी क्योंकि वहाँ सोने का महल था। जब कौआ सोकर जागा तब उसने सोने की खिड़की में से (झाँककर) कहा-अरे लड़की! तुम आ गयी हो, ठहरो, मैं तुम्हारे लिए सीढ़ी उतारता हूँ, तो कहो सोने से बनी (सोने की), चाँदी से बनी (चाँदी की) अथवा ताँबे से बनी (ताँबे की)? लड़की बोली-मैं गरीब माता की बेटी हूँ। ताँबे की बनी सीढ़ी से ही आऊँगी। लेकिन सोने की सीढ़ी से वह सोने के महल में पहुँची।

संस्कत-व्याख्याः

सन्दर्भः – प्रस्तुतगद्यांश अस्माकं पाठ्यपुस्तकात् शेमुष्याः ‘स्वर्णकाकः’ पाठात् उद्धृतः। पाठोऽयं श्री पद्मशास्त्रिणा विरचितात् ‘विश्वकथाशतकम्’ इति कथासंग्रहात् संकलितो वर्तते। (प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक शेमुषी से ‘स्वर्णकाकः’ पाठ से उद्धृत है। यह पाठ श्री पद्म शास्त्री द्वारा रचित ‘विश्वकथाशतकम्’ कथा-संग्रह से सङ्कलित है।)

प्रसङ्गः – गद्यांशोऽयम् वर्णयति यत् बालिका काकेन निर्दिष्टस्थानम् अगच्छत्, तत्र च गत्वा सा वृक्षस्योपरि स्वर्णमयप्रासादं दृष्ट्वा विस्मिता जाता। (यह गद्यांश वर्णन करता है कि लड़की कौवा द्वारा बताये गये स्थान पर गई और वहाँ जाकर वह वृक्ष के ऊपर सोने के महल को देखकर आश्चर्यचकित हो गई।)

व्याख्याः – सा बालिका भानूदयात् प्राग् एव काकेन निर्दिष्ट स्थानं आगच्छत् । वृक्षस्य उपरि दृष्ट्वा सा विस्मिता अभवत् यत् तत्र हेममयं प्रासादम् आसीत्। यदा काकः शयनं कृत्वा जागरणम् अकरोत् तदा स हेममयात् वातायनात् अकथयत्-हे बाले! त्वम् आगतवती, स्थिरा भव अहं तुभ्यं सोपानम् अवतीर्णं करोमि, कथय तु हेममयं सोपानं रजतमयं ताम्रमयं वा? बालिका अवदत् यदहं दरिद्रायाः जनन्याः पुत्री अस्मि, ताम्रमयेन एव सोपानेन आगमिष्यामि। परञ्च सा हेममयेन सोपानेन हेममयं प्रासादं प्राप्तम् अकरोत्। (वह लड़की सूरज उगने से पहले ही कौआ द्वारा बताये हुए स्थान पर पहुँच गई।

पेड़ के ऊपर देखकर वह आश्चर्यचकित हो गई कि वहाँ सोने का महल था। जब कौआ सोकर जागा तब उसने सोने की खिड़की से कहा-‘हे बालिका तुम आ गई, ठहरो, मैं तुम्हारे लिए सीढ़ी उतारता हूँ, कहो तो सोने की, चाँदी की या ताँबे की?” लड़की बोली कि “मैं गरीब माँ की बेटी हूँ, ताँबे की सीढ़ी से ही आ जाऊँगी। परन्तु वह सोने की सीढ़ी से ही सोने के महल पर पहुँची।)

अवबोधन कार्यम्

प्रश्न 1.
एकपदेन उत्तरत – (एक शब्द में उत्तर दीजिए-)
(क) बालिका तत्र कदा उपस्थिता? (बालिका कहाँ कब उपस्थित हो गई?)
(ख) काकस्य प्रासादः कुत्र आसीत् ? (कौवे का महल कहाँ था?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत – (पूरे वाक्य में उत्तर दीजिए-)
(क) बालिका कस्मात् आश्चर्यचकिता सञ्जाता? (बालिका किसलिए आश्चर्यचकित हो गई?)
(ख) कन्या काकं किम् अवदत्? (कन्या ने कौवे से क्या कहा?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) ‘विलोक्य’ इति पदे प्रत्ययं लिखत। (‘विलोक्य’ पद में प्रत्यय लिखिए।)
(ख) ‘स्वर्णमयः प्रासादः वर्तते’ एतेषु वाक्येषु विशेषणं विशेष्यं च पदम् निर्दिशत।
(यहाँ विशेषण और विशेष्य पदों का निर्देश कीजिए।)
उत्तराणि :
(1) (क) सूर्योदयात्पूर्वम् (सूर्य निकलने से पहले)।
(ख) वृक्षस्योपरि। (वृक्ष के ऊपर)।

(2) (क) वृक्षस्योपरि स्वर्णमयं प्रासादम् अवलोक्य आश्चर्यचकिता सञ्जाता। (वृक्ष के ऊपर सोने के महल को
देखकर आश्चर्यचकित हो गई।)
(ख) अहं निर्धनमातुः दुहिता अस्मि ताम्र सोपानेनैव आगमिष्यामि। (मैं गरीब माँ की बेटी हूँ। ताँबे की सीढ़ी से ही आ जाऊँगी।)

(3) (क) ल्यप्।
(ख) विशेषणम्-स्वर्णमयः, विशेष्यं च-प्रासादः।

3. चिरकालं भवने चित्रविचित्रवस्तूनि सज्जितानि दृष्ट्वा सा विस्मयं गता। श्रान्तां तां विलोक्य काकः अवदत्-“पूर्वं लघुप्रातराशः क्रियताम्-वद त्वं स्वर्णस्थाल्यां भोजनं करिष्यसि किं वा रजतस्थाल्याम् उत ताम्रस्थाल्याम्”? बालिका अवदत्- ताम्रस्थाल्याम् एव अहं – “निर्धना भोजनं करिष्यामि।” तदा सा आश्चर्यचकिता सञ्जाता यदा स्वर्णकाकेन स्वर्णस्थाल्यां भोजनं “परिवेषितम्।” न एतादृशम् स्वादु भोजनमद्यावधि बालिका खादितवती। काकोऽवदत्- बालिके! अहमिच्छामि यत् त्वम् सर्वदा अत्रैव तिष्ठ परं तव माता तु एकाकिनी वर्तते। अत: “त्वं शीघ्रमेव स्वगृहं गच्छ।”

शब्दार्थाः – चिरकालम् = बहुकालं यावत् (बहुत देर तक), भवने = प्रासादे (महल में), चित्रविचित्रवस्तूनि = अद्भुतानि वस्तूनि (अद्भुत वस्तुओं को), सज्जितानि = अलंकृतानि (सजी हुई), दृष्ट्वा = विलोक्य (देखकर), सा = (वह), विस्मितम् = आश्चर्यचकितम् (आश्चर्यचकित), गता = सञ्जाता (हुई), श्रान्ताम् = क्लान्तां (थकी हुई), ताम् = (उसे), विलोक्य = दृष्ट्वा (देखकर), काकः = वायसः (कौआ),

अवदत् = प्राह (बोला), पूर्वम् = प्राग् (पहले), लघुप्रातराशः = लघुकल्यवर्तः (थोड़ा सुबह का नाश्ता), क्रियताम् = करोतु (करो), वद = कथय (बोलो), त्वम् = (तुम), स्वर्णस्थाल्याम् = स्वर्णेन निर्मितस्थाल्यां (सोने की थाली में), भोजनम् = अशनम् (भोजन), करिष्यसि = (करोगी), किं वा = अथवा (अथवा), रजतस्थाल्याम् = रूप्येन निर्मितस्थाल्यां (चाँदी की थाली में), अकथयत् = व्याजहार (कहा), ताम्रस्थाल्याम् = लोहितायसेन निर्मितस्थाल्यां (ताँबे की थाली में), एव = निश्चयेन (ही), अहम् = मैं, निर्धना = दरिद्रा (निर्धन), भोजनम् = अशनम् (भोजन), करिष्यामि = (करूँगी), तदा = तस्मिन् समये (तब), सा = वह (लड़की), आश्चर्यचकिता = विस्मिता (आश्चर्यचकित),

सजाता = गता (हो गई), यदा = यस्मिन् काले (जब), स्वर्णकाकेन = हेमवायसेन (सुनहले कौए ने), स्वर्णस्थाल्याम् = स्वर्णेन निर्मितस्थाल्यां (सोने की थाली में), भोजनम् = अशनम् (भोजन), परिवेषितम् = पर्यवेषणं कृतम् (परोसा), न = मा (नहीं), एतादृशम् = ईदृशं (ऐसा), स्वादु = स्वादिष्टम् (स्वादिष्ट), भोजनम् = अशनं (भोजन), अद्यावधि = वर्तमानं दिनं यावत् (आज तक), बालिका = सा कन्या (उस लड़की ने), खादितवती = भक्षितवती (खाया था),

काकः = वायसः (कौआ), अवदत् = ब्रूते (बोला), बालिके = कन्ये (बालिका), अहमिच्छामि = अहमभिलषामि (मैं चाहता हूँ), यता = कि, त्वम् = (तुम), सर्वदा = सर्वस्मिन् काले (हमेशा), अत्रैव = एतस्मिन् स्थाने एव (यहीं), तिष्ठ = वासं कुरु (रहो), परम् = परञ्च (लेकिन), तव = ते (तुम्हारी), माता = जननी (माता), वर्तते = अस्ति (है), एकाकिनी = एकला (अकेली), वम् = (तुम), शीघ्रमेव = द्रुतमेव (शीघ्र ही), स्वगृहम् = स्वगेहं (अपने घर), गच्छ = गमनं कुरु (जाओ)।

हिन्दी-अनुवादः

सन्दर्भ – प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक ‘शेमुषी’ के ‘स्वर्णकाकः’ नामक पाठ से उद्धृत है। यह पाठ श्री . । . पद्मशास्त्री द्वारा रचित ‘विश्वकथाशतकम्’ नामक कथा-संग्रह से संकलित है।।

प्रसंग – यहाँ वर्णन है कि सोने के महल में सुनहले कौए ने बालिका के साथ कैसा व्यवहार किया।

हिन्दी-अनुवाद – बहुत देर तक महल में अत्यन्त अदभुत वस्तुओं को सजी हुई देखकर वह आश्चर्यचकित हो गयी। उसे थकी हुई देखकर कौआ उससे बोला कि पहले आप थोड़ा-सा सुबह का नाश्ता कर लो, बोलो तो सोने की थाली में भोजन करोगी अथवा चाँदी की थाली में या ताँबे की थाली में ? लड़की बोली कि मैं गरीब हूँ, मैं तो ताँबे की थाली में ही भोजन करूँगी। तब वह आश्चर्यचकित हो गयी जब सुनहले कौए ने उसे सोने की थाली में भोजन परोसा। लड़की ने ऐसा स्वादिष्ट भोजन आज तक नहीं खाया था। कौआ उससे बोला कि मैं चाहता हूँ कि तुम हमेशा यहीं रहो लेकिन तुम्हारी माता अकेली हैं। अतः तुम जल्दी ही अपने घर जाओ।

संस्कत-व्याख्याः

सन्दर्भ: – प्रस्तुतः गद्यांशः अस्माकं पाठ्यपुस्तकात् शेमुष्याः ‘स्वर्णकाकः’ पाठात् उद्धृतः। पाठोऽयं श्रीपद्मशास्त्रिणा विरचितात् ‘विश्वकथाशतकम्’ इति कथासंग्रहात् सङ्कलितो वर्तते। (प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक ‘शेमुषी’ के स्वर्णकाकः’ पाठ से उद्धृत है। यह पाठ श्रीपद्मशास्त्री-रचित ‘विश्वकथाशतकम्’ से संकलित है।)

प्रसंग: – वर्णितमत्र यत् स्वर्णकाक: स्वप्रासादे बालिकया सह कीदृशं व्यवहारम् अकरोत्। (इसमें बालिका के साथ कौआ के सद्व्यवहार का वर्णन किया गया है।)

व्याख्या: – बहकालं यावत् स्वर्णप्रासादे अलङ्कृतानि अति अद्भुतवस्तुनि विलोक्य सा आश्चर्यचकिता अभवत्। तां क्लान्तां दृष्ट्वा काकः तामवदत् यत् पूर्वं भवती लघुकल्यवर्तं करोतु, कथय तु स्वर्णमयस्थाल्यां, रजतमयस्थाल्यां ताम्रमयस्थाल्यां वा भोजनं करिष्यसि। बालिका अवदत् यदहं निर्धना अस्मि, अहं तु ताम्रमयस्थाल्यामेव अशनं करिष्यामि। यदा स्वर्णकाकः तस्यै स्वर्णमयस्थाल्यां भोजनम् अददात्। तदा सा चकिता अभवत्। बालिका ईदृशं स्वादिष्टभोजनं वर्तमानं दिनं यावत् न अखादत्। काकः तामवदत् यदहम् अभिलषामि यत् त्वम् सदा अत्रैव वासं कुरु परं तव माता एकला अस्ति। अतः त्वं शीघ्रम् एव स्वगृहं गच्छ।

(बहुत देर तक सोने के महल में सजी हुई अनोखी वस्तुओं को देखकर वह आश्चर्यचकित हो गई। उसे थका हुआ देखकर उससे कौआ बोला कि पहले आप थोड़ा कलेवा कर लो, बोलो तुम सोने की थाली में, चाँदी की थाली में या ताँबे की थाली में भोजन करोगी। बालिका बोली कि मैं गरीब हूँ, मैं तो ताँबे की थाली में ही खाना खाऊँगी (भोजन करूँगी)। जब सोने का कौवा उसके लिए सोने की थाली में भोजन लाया तो वह चकित हो गई। लड़की ने ऐसा स्वादिष्ट भोजन आज तक नहीं खाया। कौआ उससे बोला कि मैं चाहता हूँ कि तुम सदा यहाँ ही रहो परन्तु तुम्हारी माता अकेली हैं। अतः तुम शीघ्र ही अपने घर चली जाओ।)

अवबोधन कार्यम्

प्रश्न 1.
एकपदेन उत्तरत- (एक शब्द में उत्तर दीजिए-)
(क) कानि दृष्ट्वा बालिका विस्मयं गता? (किन्हें देखकर बालिका आश्चर्य करने लगी?) ।
(ख) स्वर्णकाकेन तस्मै कस्यां भोजनं पर्यवेषितम् ? (सोने के कौवे ने कैसी थाली में उसे भोजन परोसा?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत- (पूरे वाक्य में उत्तर दीजिए-)
(क) काक: बालिका कस्मात् न वारयति स्म? (कौवे ने बालिका को क्यों नहीं रोका?)
(ख) श्रान्तां बालिकाम् अवलोक्य काकः किं कर्तुम् अवदत् ? (थकी बालिका को देखकर कौवे ने क्या करने के लिए कहा?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) ‘दृष्ट्वा सा विस्मयं गता’ अत्र सा इति सर्वनाम पदं कस्मै प्रयुक्तम्? (‘दृष्ट्वा सा विस्मयं गता’ यहाँ ‘सा’ सर्वनाम पद किसके लिए प्रयोग हुआ है?)
(ख) ‘त्वं शीघ्रमेव स्वगृहं गच्छ’ इति कः कम् अकथयत् ? (‘तुम शीघ्र अपने घर जाओ’ यह किसने किससे
कहा?)
उत्तराणि :
(1) (क) चित्रविचित्र वस्तूनि (रंग-बिरंगी अद्भुत वस्तुओं को)।
(ख) स्वर्णस्थाल्याम् (सोने की थाली में)।

(2) (क) यतः तस्य माता गृहे एकाकिनी अस्ति। (क्योंकि उसकी माँ घर पर अकेली है।) श्रान्तां बालिका अवलोक्य काकः लघुप्रातराशं कर्तुम् अवदत्। (थकी बालिका को देखकर कौवे ने थोड़ा कलेवा करने के लिए कहा।)

(3) (क) बालिकायै (बालिका के लिए।)
(ख) काकः बालिकाम् अकथयत्। (कौवे ने बालिका से कहा।)

4. इत्युक्त्वा काकः कक्षाभ्यन्तरात् तिम्रः मञ्जूषाः निस्सार्य तां प्रत्यवदत्- “बालिके! यथेच्छं गृहाण मञ्जूषामेकाम्।” लघुतमा मञ्जूषां प्रगृह्य बालिकया कथितम् इयत् एत्र मदीयतण्डुलानां मूल्यम्।
गृहमागत्य तया मञ्जूषा समुद्घाटिता, तस्यां महार्हाणि हीरकाणि विलोक्य सा प्रहर्षिता तद्दिनाद्धनिका च सजाता।

शब्दार्थाः – इत्युक्त्वा = एवं कथयित्वा (ऐसा कहकर), काकः = वायसः (कौए ने), कक्षाभ्यन्तरात् = कोष्ठस्य गर्भात् (कमरे के अन्दर से), तिस्त्रः = तीन, मञ्जूषाः = पेटिकाः (पेटियाँ), निस्सार्य = निष्कासनं कृत्वा (निकालकर), ताम् = उससे, लड़की से, प्रत्यवदत् = अकथयत् (कहा), बालिके = कन्ये (बाला), यथेच्छम् = इच्छानुसारेण (अपनी इच्छा के अनुसार), गृहाण = ग्रहणं कुरु (ले लो), मञ्जूषामेकाम् = एकां पेटिकां (एक पेटी), लघुतमाम् = क्षुद्रतमा (सबसे स्वकाकः छोटी), मञ्जूषाम् = पेटिकां (पेटी को),

प्रगृह्य = ग्रहणं कृत्वा (लेकर), बालिकया = कन्यया (बालिका द्वारा), कथितम् = उक्तम् (कहा गया), इयत् एव = एतन्मात्रम् एव (इतना ही), मदीय = मम (मेरे), तण्डुलानाम् = अक्षतानां (चावलों का), मूल्यम् = अर्घम् (कीमत/मूल्य), गृहमागत्य = सदनमागत्य (घर आकर), तया = (उसके द्वारा), मञ्जूषा = पेटिका (पेटी), समुद्घाटिता = भेदनं कृतं (खोली गई), तस्याम् = (उसमें), महार्हाणि = बहुमूल्यानि (बहुमूल्य), हीरकाणि – रत्नमुख्यानि (हीरों को), विलोक्य = दृष्ट्वा (देखकर), सा = (वह), प्रहर्षिता = आनन्दिता अभवत् (आनन्दित हुई), तदिनाद्धनिका = तस्मात् वासरात् धनिका (उसी दिन से धनवती), सजाता = अभवत् (हो गई)।

हिन्दी-अनुवादः

सन्दर्भ – प्रस्तुत गद्यांश हमारी पाठ्य-पुस्तक ‘शेमुषी’ के ‘स्वर्णकाकः’ नामक पाठ से उद्धृत है। यह पाठ श्री। पद्मशास्त्री द्वारा रचित ‘विश्वकथाशतकम्’ नामक कथा-संग्रह से संकलित है।

प्रसंग – यहाँ लड़की के लोभरहित होने तथा कौए की उसके प्रति उदारता का वर्णन है।

हिन्दी-अनुवाद – ऐसा कहकर कौए ने कमरे के अन्दर से तीन पेटियाँ (सन्दूकें) निकालकर उससे कहा-बालिका! अपनी इच्छानुसार एक सन्दूक ले लो। सबसे छोटी सन्दूक लेकर लड़की ने कहा-यह ही मेरे चावलों का मूल्य है। घर आकर उसने सन्दूक खोला, उसमें (सन्दूक में) बहुमूल्य हीरों को देखकर वह प्रसन्न हो गयी और उसी दिन से धनिक हो गई।

संस्कत-व्यारव्याः

सन्दर्भ: – प्रस्तुतगद्यांशः अस्माकं पाठ्यपुस्तकात् शेमुष्या: ‘स्वर्णकाकः’ पाठात् उद्धृतः। पाठोऽयं श्रीपद्मशास्त्रिणा विरचितात् ‘विश्वकथाशतकम् इति कथासंग्रहात् संकलितो वर्तते। (प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक ‘शेमुषी’ से ‘स्वर्णकाकः’ पाठ से लिया गया है। यह पाठ श्री पद्मशास्त्री-रचित ‘विश्वकथाशतकम्’ कथा-संग्रह से संकलित है।)

प्रसंग: – गद्यांशोऽयं वर्णयति बालिकायाः लोभराहित्यं काकस्य च तां प्रति औदार्यम्। (यह गद्यांश बालिका की । लोभहीनता तथा कौए की उसके प्रति उदारता का वर्णन करता है।)

व्याख्या: – एवं कथयित्वा काकः कोष्ठस्य गर्भात् तिस्रः पेटिकाः बहिः आनीय तां (बालिकाम्) अवदत्-हे बालिके! स्वेच्छया एकां पेटिकां गृहाण। बालिका क्षुद्रतमां पेटिकां गृहीत्वा अवदत्-एतन्मात्रम् मम अक्षतानां अर्घम् अस्ति। गृहम् आगत्य सा पेटिकाम् उद्घाटयत्। तस्यां बहुमूल्यानि हीरकाणि दृष्ट्वा सा प्रसन्ना अभवत्। तस्मात् वासरात् च धनवती अभवत्। (इस प्रकार कहकर कौआ ने कोटर से तीन पेटियाँ लाकर उस बालिका से कहा-हे बालिके! अपनी इच्छा से एक पेटी ले लो। बालिका ने सबसे छोटी पेटी को लेकर कहा-‘इतना ही मेरे चावलों का मूल्य है।’ घर आकर उसने पेटी को खोला। उसमें बहुमूल्य हीरों को देखकर वह प्रसन्न हो गई। उसी दिन से वह धनवाली (धनाढ्य) हो गई।)

अवबोधन कार्यम्

प्रश्न 1.
एकपदेन उत्तरत- (एक शब्द में उत्तर दीजिए-)
(क) काकः कति मञ्जूषाः आनयत? (कौवा कितनी पेटियाँ लाया?)
(ख) बालिका कीदृशी मञ्जूषां गृहीतवती? (बालिका ने कैसी पेटी ली?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत- (पूरे वाक्य में उत्तर दीजिए-)
(क) मञ्जूषायां कानि आसीत्? (कितनी मंजूषाएँ थीं ?)
(ख) मञ्जूषां प्रगृह्य बालिका किमवदत् ? (पेटी को पाकर लड़की ने क्या कहा?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) ‘तां प्रत्यवदत्’ अत्र ‘ताम्’ इति सर्वनाम पदं कस्य स्थाने प्रयुक्तम्? (तां प्रत्यवदत् यहाँ ‘ताम्’ सर्वनाम किसके लिए प्रयोग हुआ है?)
(ख) गद्यांशात् ‘पेटिका’ इति पदस्य पर्यायं चित्वा लिखत। (गद्यांश से ‘पेटिका’ का पर्यायवाची लिखिए।)
उत्तराणि :
(1) (क) तिस्रः (तीन)।
(ख) लघुतमाम् (सबसे छोटी को)।

(2) (क) मञ्जूषायां बहुमूल्यानि हीरकाणि आसन्। (मंजूषा में महँगे हीरे थे।)
(ख) तया कथतम् इयम् मे तण्डुलानां मूल्यम्। (यह मेरे चावलों की कीमत है, ऐसा उसने कहा।)

(3) (क) बालिकाम् (बालिका को)।
(ख) मञ्जूषा (पेटी)।

5. तस्मिन्नेव ग्रामे एका अपरा लुब्धा वृद्धा न्यवसत्। तस्या अपि एका पुत्री आसीत्। ईर्ष्णया सा तस्य स्वर्णकाकस्य रहस्यम् ज्ञातवती। सूर्यातपे तण्डुलान् निक्षिप्य तयापि स्वसुता रक्षार्थं नियुक्ता। तथैव स्वर्णपक्षः काकः तण्डुलान् भक्षयन् तामपि तत्रैवाकारयत्। प्रातस्तत्र गत्वा सा काकं निर्भयन्ती प्रावोचत्-“भो नीचकाक! अहमागता, मां तण्डुलमूल्यं प्रयच्छ।” काकोऽब्रवीत्-“अहं त्वत्कृते सोपानम् अवतारयामि। तत्कथय स्वर्णमयं रजतमयं ताम्रमयं वा।” गर्वितया बालिकया प्रोक्तम्-“स्वर्णमयेन सोपानेन अहम् आगच्छामि।” परं स्वर्णकाकस्तत्कृते ताम्रमयं सोपानमेव प्रायच्छत्। स्वर्णकाकस्तां भोजनमपि ताम्रभाजने एव अकारयत्।

शब्दार्थाः – तस्मिन् एव ग्रामे = अमुष्मिन् एव वसथे (उसी गाँव में), एकाऽपरा = एका अन्या (एक दूसरी), लुब्धा = लोभवशीभूता (लोभी), वृद्धा = जरठा (बुढ़िया), न्यवसत् = वासं करोति स्म (रहती थी), तस्या = अमुष्याः (उसकी), अपि = (भी), एका पुत्री = एका दुहिता (एक पुत्री), आसीत् = अवर्तत (थी), ईर्ष्णया = मत्सरेण (ईर्ष्या से), सा = (वह), तस्य स्वर्णकाकस्य = अमुष्य हेमवायसस्य (उस सोने के कौए के), रहस्यम् = गुप्तभेदं (रहस्य को), ज्ञातवती = बोधमाना (जान गई), सूर्यातपे = भानोः उष्णतायां (धूप में), तण्डुलान् = अक्षतान् (चावलों को),

निक्षिप्य = स्थिरीकृत्य (रखकर), तयापि = (उसने भी), स्वसुता = स्वपुत्री (अपनी पुत्री), रक्षार्थम् = रक्षायाः कृते (रक्षा के लिए), नियुक्ता = नियुक्ताम् अकरोत् (नियुक्त किया), तथैव = तेनैव प्रकारेण (उसी तरह), स्वर्णपक्षः काकः = हेमपक्षः वायसः (सुनहले पंखों वाले कौए ने), तण्डुलान् = अक्षतान् (चावलों को), भक्षयन् = खादन् (खाते हुए), तत्रैव = तस्मिन् एव स्थाने (वहीं), आकारयत् = आगन्तुम् अकथयत् (बुलाया), प्रातस्तत्र = प्रातःकालं तत्र (सुबह वहाँ), गत्वा = प्रस्थानं कृत्वा (जाकर), सा = (वह),

काकम् = वायसं (कौए की), निर्भसंयन्ती = निन्दां कुर्वन्ती (निन्दा करती हुई), प्रावोचत् = अवदत् (बोली), भो नीचकाकः = हे क्षुद्रवायसः (अरे नीच कौए), अहमागता = अहम् अत्र आगच्छम् (मैं यहाँ आ गयी हूँ), मह्यम् = मम कृते (मुझे), तण्डुलमूल्यम् = अक्षतानां मूल्यं (चावलों का मूल्य), प्रयच्छ = यच्छ (दो), काकः अब्रवीत् = वायसः अवदत् (कौआ बोला), अहम् = मैं, त्वत्कृते = तुभ्यम् (तुम्हारे लिए), सोपानम् अवतारयामि = सोपानम् अवतीर्णं करोमि (सीढ़ी उतारता हूँ।), तत्कथय = तु वद (तो बोलो), स्वर्णमयम् = हेममयं (सोने की बनी), रजतमयम् = रूप्यमयं (चाँदी की बनी),

ताम्रमयम् = लोहितायसमयं (ताँबे की बनी), वा = अथवा, गर्वितया = अभिमानेन (घमण्ड से), बालिकया = कन्यया (लड़की), प्रोक्तम् = अब्रवीत् (बोली), स्वर्णमयेन सोपानेन = हेमनिर्मितसोपानेन (सोने की बनी सीढ़ी से), अहम् आगच्छामि = अहम् आगमिष्यामि (मैं आऊँगी), परम् = परञ्च (लेकिन), स्वर्णकाकः = हेमवायसः (सुनहले कौए ने), तत्कृते = तस्यै (उसके लिए), ताम्रमयं सोपानमेव = लोहितायसमयं सोपानम् एव (ताँबे की सीढ़ी ही), प्रायच्छत् = अददात् (दी), स्वर्णकाकः = हेमवायसः (सुनहले कौए ने), ताम् = (उसको), भोजनमपि = अशनमपि (भोजन भी), ताम्रभाजने = लोहितायसपात्रे (ताँबे के बर्तन में), एव अकारयत् = एव पर्यवेषयत् (ही कराया/ही दिया)।

हिन्दी-अनुवादः

सन्दर्भ – प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक ‘शेमुषी’ के ‘स्वर्णकाकः’ पाठ से उद्धृत है। यह पाठ श्रीपद्मशास्त्री द्वारा रचित ‘विश्वकथाशतकम्’ नामक कथा-संग्रह से संकलित है।

प्रसंग – प्रस्तुत गद्यांश में वर्णन है कि एक लालंची महिला ने अपनी पुत्री से धूप में रखे चावलों की रक्षा करने को कहा। साथ ही सुनहले कौए के चावल खाने पर उससे उनका मूल्य लेने को भी कहा।

हिन्दी-अनुवाद – उसी गाँव में एक दूसरी लालची वृद्धा (बूढ़ी) औरत रहती थी। उसकी भी एक बेटी थी। ईर्ष्या से वह उस सुनहले कौए के रहस्य को जान गई। उसने भी धूप में चावलों को रखकर अपनी पुत्री को (उनकी) रक्षा करने के लिए नियुक्त किया। उसी तरह सुनहले पंखों वाले कौए ने चावलों को खाते हुए, उसे भी वहीं बुलाया। सुबह वहाँ जाकर वह कौए की निन्दा करती हुई बोली-अरे नीच कौए! मैं आ गयी, मुझे चावलों का मूल्य दो। कौआ बोला-मैं तुम्हारे लिए सीढ़ी उतारता हूँ, तो बोलो सोने की बनी, चाँदी की बनी अथवा ताँबे की बनी। घमण्ड से लड़की बोली-मैं सोने की बनी सीढ़ी से आऊँगी। लेकिन सुनहले कौए ने उसके लिए ताँबे की सीढ़ी ही दी। सुनहले कौए ने उसको भोजन भी ताँबे के बर्तन में ही कराया।

संस्कत-व्याख्याः

सन्दर्भ: – गद्यांशोऽयम् अस्माकं पाठ्यपुस्तकात् शेमुष्याः ‘स्वर्णकाकः’ पाठात् उद्धृतः। पाठोऽयं श्रीपद्मशास्त्रिणा विरचितात् ‘विश्वकथाशतकम्’ इति कथासंग्रहात् सङ्कलितो वर्तते। (प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक ‘शेमुषी’ से ‘स्वर्णकाकः’ पाठ से लिया गया है। यह पाठ श्री पद्मशास्त्री रचित ‘विश्वकथाशतकम्’ से संकलित है।)

प्रसङ्गः – वर्णितमत्र वर्तते यदेका लुब्धा स्त्री सूर्यातपे अक्षतान् निक्षिप्य स्वपुत्रीं तेषां रक्षार्थं नियुक्तवती। स्वर्णकाके च अक्षतान् खादिते तेषां मूल्यं ग्रहीतुम् अकथयत्। (इसमें वर्णन किया गया है कि एक लोभी स्त्री धूप में चावल रखकर अपनी बेटी को उनकी रखवाली के लिए नियुक्त करती है। स्वर्णकाक चावलों को खाता है और वह उनका मूल्य ग्रहण करने के लिए कहती है।)

व्याख्या: – तस्मिन् एव ग्रामे एका अन्या लोभान्विता वृद्धा स्त्री वसति स्म। तस्याः अपि एका दुहिता आसीत्। मत्सरेण सा वृद्धा तस्य हेमवायसस्य रहस्यम् अजानीत्। सा अपि सूर्यातपे अक्षतान् स्थिरीकृत्य स्वपुर्वी तेषां रक्षार्थं नियुक्ताम् अकरोत्। तथैव हेमवायसः तण्डुलान् खादन् तामपि तत्रैव आगन्तुम् अकथयत् सा (बालिका) प्रातः तत्र प्रस्थानं कृत्वा काकस्य निन्दा कुर्वन्ती अवदत् यद् भो अधमकाक! अहमत्र उपस्थिता अस्मि। मम कृते अक्षतानां मूल्यं देहि। काकः अवदत् यत् अहं तुभ्यं सोपानम् अवतारयामि, कथय तावत् स्वर्णमयं, रूप्यमयं ताम्रमयं वा।

गर्वयुक्ता सा बालिका प्रत्यवदत् यदहं स्वर्णमयेन सोपानेन आगमिष्यामि। परं स्वर्णकाकः तस्यै ताम्रसोपानमेव अददात्। स्वर्णकाकः तस्यै अशनमपि ताम्रपात्रे एव पर्यवेषयत्। (उसी गाँव में एक दूसरी लोभी बुढ़िया रहती थी। ईर्ष्या के कारण वह बुढ़िया स्वर्णकाक के रहस्य को जानती थी। उसने भी धूप में चावल रखकर अपनी बेटी को रखवाली के लिए नियुक्त कर दिया। उसी प्रकार सोने का कौवा चावलों को खाते हुए उसे भी आने के लिए कह गया।

वह लड़की भी प्रातः वहाँ प्रस्थान करके कौवे की निन्दा करती हुई बोली-अरे नीच कौवे; मैं यहाँ उपस्थित हो गई हूँ। मेरे लिए चावलों का मूल्य दो। कौआ बोला कि मैं तेरे लिए सीढ़ी उतारता हूँ, बोलो तो सोने की चाँदी की अथवा ताँबे की? गर्वयुक्त उस लड़की ने उत्तर दिया कि मैं सोने की सीढ़ी से ही आऊँगी। लेकिन स्वर्णकाक ने उसके लिए ताँबे की सीढ़ी दी। स्वर्ण काक ने उसके लिए भोजन भी ताँबे की थाली में दिया।)

अवबोधन कार्यम

प्रश्न 1.
एकपदेन उत्तरत- (एक शब्द में उत्तर दीजिए-)
(क) निर्धारित स्थान प्राप्य लुब्धायाः वृद्धाया पुत्री काकं केन शब्देन समबोधयत्।
(निर्धारित स्थान पर पहुँच कर लोभी वृद्धा की बेटी ने कौवे को किस शब्द से संबोधित किया?)
(ख) सा कस्य रहस्यं न ज्ञातवती? (वह किसके रहस्य को नहीं जान सकी?)

प्रश्न 2. पूर्णवाक्येन उत्तरत- (पूरे वाक्य में उत्तर दीजिए-)
(क) बालिका स्वर्णकाकस्य रहस्यं कस्मात् न ज्ञातवती? (बालिका स्वर्णकाक के रहस्य को क्यों नहीं जान सकी?)
(ख) बालिका निर्भत्सयन्ती काकं किं प्रावोचत्? (बालिका ने निन्दा करते हुए कौवा को क्या कहा?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) ‘पुत्री’ इति पदस्य पर्यायवाचि पदं गद्यांशात् अन्विष्य लिखत।
(‘पुत्री’ पद का पर्यायवाची पद गद्यांश से ढूँढ़ कर लिखिए।)
(ख) ‘तस्मिन्नेव ग्रामे’ एतयोः पदयोः किं विशेषणपदम् ?
(तस्मिन्नेव ग्रामे’ इनमें विशेषण पद कौनसा है?)
उत्तराणि :
(1) (क) नीच काक ! (नीच कौआ।)।
(ख) स्वर्णकाकस्य। (सोने के कौवे का)।

(2) (क) ईय॑या बालिका स्वर्णकाकस्य रहस्यं न ज्ञातवती। (ईर्ष्या के कारण बालिका स्वर्णकाक के रहस्य को न जान सकी।)
(ख) सा प्रांवोचत्-‘भो नीचकाक ! अहम् आगता मह्यं तण्डुल मूल्यं प्रदत्य। (वह बोली- अरे नीच कौवे! मैं आ गई हूँ, मेरे चावलों का मूल्य दो।)

(3) (क) सुता (बेटी)।
(ख) तस्मिन् (उसमें)।

6. प्रतिनिवृत्तिकाले स्वर्णकाकेन कक्षाभ्यन्तरात् तिम्रः मञ्जूषाः तत्पुरः समुत्क्षिप्ताः। लोभाविष्टा सा बृहत्तमा मञ्जूषां गृहीतवती। गृहमागत्य सा तर्षिता यावद् मञ्जूषामुद्घाटयति तावत् तस्यां भीषणः कृष्णसर्या विलोकितः। लुब्धया बालिकया लोभस्य फलं प्राप्तम्। तदनन्तरं सा लोभं पर्यत्यजत्।

शब्दार्था: – प्रतिनिवत्तिकाले = प्रत्यागमनस्य समये (वापस लौटते समय), स्वर्णकाकेन = हेमवायसेन (सोने के कौए द्वारा), कक्षाभ्यन्तरात् = कोष्ठस्य गर्भात् (कमरे के अन्दर से), तिस्रः = तीन, मञ्जूषाः = पेटिकाः (सन्दूकें/पेटियाँ), तत्पुरः = तस्याः अग्रे (उसके सामने), समुत्क्षिप्ताः = निक्षिप्ताः (रखी गईं), लोभाविष्टा = लोभवशीभूता (लोभ से घिरी), सा = (उसने), बृहत्तमाम् = विशालतमा (सबसे बड़ी), मञ्जूषाम् = पेटिकां (सन्दूक/पेटी), गृहीतवती = ग्रहणम् अकरोत् (ग्रहण की/ली),

गृहमागत्य = गेहम् आगत्य (घर आकर), सा = (उसने), हर्षिता = प्रसन्ना भवन्ती (प्रसन्न होती हुई), यावद् = यस्मिन्नेव क्षणे (जैसे ही), मञ्जूषामुद्घाटयति = पेटिकायाः भेदनम् अकरोत् (सन्दूक/पेटी को खोला), तावत् = तद्वत् एव (वैसे ही), तस्याम् = (उसमें), भीषणः = भयंकरः (भयंकर), कृष्णसर्पः = श्यामः भुजगः (काले साँप को), विलोकितः = अपश्यत् (देखा), लुब्धया बालिकया = लोभवशीभूतया बालिकया (लालची लड़की को), लोभस्य फलम् = परद्रव्याभिलाषस्य परिणाम (लोभ का फल), प्राप्तम् = (प्राप्त हो गया), तदनन्तरम् = तत्समयपश्चात् (उसके पश्चात्), सा = (उसने), लोभम् = परद्रव्याभिलाषां (लोभ को/लालच को), पर्यत्यजत् = अत्यजत् (छोड़ दिया)।

हिन्दी-अनुवादः

सन्दर्भ – प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक ‘शेमुषी’ के ‘स्वर्णकाकः’ पाठ से उद्धृत है। यह पाठ श्रीपद्मशास्त्री द्वारा रचित ‘विश्वकथाशतकम्’ कथा-संग्रह से संकलित है।

प्रसंग – प्रस्तुत गद्यांश में लड़की द्वारा कौए से सबसे बड़ी सन्दूक लेने, लोभ का फल प्राप्त करने तथा लालच का त्याग करने का वर्णन है।

हिन्दी-अनवाद – (घर) वापस लौटने के समय सनहले कौए ने कमरे के अन्दर से (लाकर) तीन पेटियाँ बालिका के सामने रखी। लोभ से घिरी लड़की ने सबसे बड़ी सन्दूक ग्रहण की। घर आकर उसने प्रसन्न होते हुए जैसे ही संदूक को खोला वैसे ही उसमें भयंकर काले साँप को देखा। इस प्रकार उस लड़की को लोभ का फल प्राप्त हो गया। उसके पश्चात् उसने लोभ छोड़ दिया।

संस्कत-व्यारव्याः

सन्दर्भ: – प्रस्तुतगद्यांशः अस्माकं पाठ्यपुस्तकात् शेमुष्याः ‘स्वर्णकाकः’ पाठात् उद्धृतः। पाठोऽयं श्रीपद्मशास्त्रिणा विरचितात् ‘विश्वकथाशतकम्’ इति कथासंग्रहात् सङ्कलितो वर्तते। (प्रस्तुत गद्यांश हमारी पाठ्यपुस्तक शेमुषी के ‘स्वर्णकाकः’ पाठ से लिया गया है। यह पाठ श्री पदम्शास्त्री रचित ‘विश्वकथाशतकम्’ कथा संग्रह से संकलित है।)

प्रसङ्गः – गद्यांशोऽयं वर्णयति यद् लुब्धा बालिका लोभस्य फलं प्राप्तम् अकरोत्। (यह गद्यांश वर्णित करता है कि लोभी बालिका लोभ का फ़ल पाती है।)

व्याख्याः – प्रत्यागमनस्य समये स्वर्णकाक: कोष्ठस्य गर्भात् तिम्रः पेटिकाः बालिकायाः समक्षं निक्षिप्तवान्। लुब्धा बालिका विशालतमा पेटिकाम् अग्रहीत्। गेहम् आगता प्रसन्ना भवन्ती सा यावदेव पेटिकाम् उद्घाटयत् तावदेव अन्तः स्थितं भयङ्करम् कृष्णसर्पम् अपश्यत्। एवं सा बालिका लोभस्य फलं प्राप्तम् अकरोत्। तत्पश्चात् सा लोभम् अत्यजत्। (लौटते समय स्वर्ण काक ने कमरे के अन्दर से लाकर तीन पेटियाँ बालिका के सामने रख दी। लोभी लड़की ने सबसे बड़ी पेटी ली। घर आई हई उसने जब पेटी को खोला तो उसके अन्दर एक भयंकर काला नाग देखा। इस प्रकार उस बालिका ने लोभ का फल प्राप्त किया। इसके बाद उसने लोभ त्याग दिया।)

अवबोधन कार्यम्

प्रश्न 1.
एकपदेन उत्तरत – (एक शब्द में उत्तर दीजिए-)
(क) लोभाविष्टा कन्या कीदृशी मञ्जूषां गृहीतवती? (लोभी कन्या ने कैसी मञ्जूषा ग्रहण की?)
(ख) लुब्धया बालिकया कस्य फलं प्राप्तम्? (लोभी बालिका ने किसका फल प्राप्त किया?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत- (पूरे वाक्य में उत्तर दीजिए-)
(क) मञ्जूषाम् उद्घाट्य सा किम् अपश्यत् ? (मञ्जूषा को खोलकर उसने क्या देखा?)
(ख) काकः तिस्त्रः मञ्जूषा कदा आनयत्? (कौआ तीन मञ्जूषाएँ कब लाया?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) ‘तदनन्तरं सा लोभं पर्यत्यजत्’ अत्र सा इति सर्वनामपदं कस्य स्थाने प्रयुक्तम्?
(यहाँ ‘सा’ सर्वनाम पद किस स्थान पर प्रयुक्त हुआ है।)
(ख) ‘कृष्णसर्पो’ अत्र विशेष्य पदं किम्? (‘कृष्णसर्पो’ यहाँ विशेष्य पद क्या है?)
उत्तराणि :
(1) (क) वृहत्तमाम्। (सबसे बड़ी को)।
(ख) लोभस्य (लोभ का)।
(2) (क) मञ्जूषाम् उद्घाट्य सा कृष्ण सर्पमपश्यत्। (मंजूषा को खोलकर उसने काला साँप देखा।)
(ख) प्रतिनिवृत्तिकाले स्वर्णकाकः तिस्रः मञ्जूषाः आनयत्। (लौटते समय स्वर्णकाक तीन पेटियाँ लाया।)
(3) (क) लुब्धा बालिकाया। (लोभी बालिका के स्थान पर।)
(ख) ‘सर्पः’ अत्र विशेष्य पदम्।

JAC Board Class 9th Social Science Notes History Chapter 5 Pastoralists in the Modern World

→  Introduction

• Pastoralism is a way of keeping animals such as cattle, sheep, camels or goats, and involves moving from one place to another in search of water and food.
• Pastoralism has been important in societies like India and Africa for years.

→ Pastoral Nomads and their Movements in the Mountains

• Nomads are people who do not live at one place, but move from one place to another to earn their living.
• The Gujjar Bakarwals of Jammu and Kashmir are great herders of goat and sheep.
• In winter, when the high mountains were covered with snow, the Gujjars lived along with their herds in the low hills of the Shiwalik range.
• By the end of April, they began their northern march for their summer grazing grounds. They crossed the Pir Panjal passes and entered the valley of Kashmir.
• By the end of September, the Bakarwals were on the move again, this time on their downward journey, back to their winter base. When the high mountains were covered with snow, the herds were grazed in the low hills.
• In a different area of the mountains, the Gaddi shepherds of Himachal Pradesh had a similar cycle of seasonal movement.
• This pattern of cyclical movement between summer and winter pastures was typical of many pastoral communities of the Himalayas, including the Bhotiyas, Sherpas and Kinnauris. All of them had to adjust to seasonal changes and make effective use of available pastures in different places.

→ On the Plateaus, Plains and Deserts

• Not all pastoralists operated in the mountains. They were also to be found in the pla-teaus, plains and deserts of India.
• Dhangars were an important pastoral community of Maharashtra. Most of them were shepherds. Some were blanket
  weavers. While others were buffalo herders.
• The Dhangars stayed in the central plateau of Maharashtra during the monsoon. Nothing but dry crops like bajra could be sown here.
• By October, the Dhangars harvested their bajra and started on their move to west. After a march of about a month, they reached the Konkan.
• With the onset of the monsoon, the Dhangars left the Konkan and the coastal areas with their flocks and returned to their settlements on the dry plateau.
• Other pastoral communities of Karnataka and Andhra Pradesh were the Gollas (herded cattle), the Kurumas and Kurubas (reared sheep and goats and sold woven blankets).
• Banjaraswere yet another well-known group of graziers. They were found in the vil-lages of Uttar Pradesh, Punjab, Rajasthan, Madhya Pradesh and Maharashtra.
• In the deserts of Rajasthan, lived the Raikas. They combined cultivation within pastoralism.
• One group of Raikas known as the Maru (desert) Raikas herded camels and another group reared sheep and goats.

→ Colonial Rule and Pastoral Life

• Under colonial rule, the life of Pastoralists changed dramatically. Their grazing grounds shrank, their movements were regulated, and the revenue they had to pay increased. Their agricultural stock declined and their trades and crafts were adversely affected.
• The colonial government wanted to transform all grazing lands into cultivated farms.
• By expanding cultivation, it could increase its revenue collection.
• Expansion of cultivation inevitably meant the decline of pastures and a problem for pastoralists.
• The Forest Acts made by the British Government changed the lives of pastoralists. Some
  forest which produced commercially valuable timber like ‘deodar’ or ‘sal’ were declared as ‘reserved’. .
• In the reserved forests, no pastoral activity was allowed, and in the protected forests, their activities were strictly restricted.
• In 1871, the British Government in India passed the Criminal Tribes Act. This act classified communities of craftsmen, traders and pastoralists as Criminal Tribes.
• As a result of this Act, these communities were expected to live only in notified vil¬lage settlements and they were not allowed to move without a permit from the British government.
• In the mid nineteenth century, Grazing tax was introduced by British Government in most pastoral lands of India.
• In order to increase income, the British Government imposed tax even on animals. The tax per head of cattle went up rapidly and the system of collection was made increasingly efficient.
• By the decade of1880s the government began collecting taxes directly from the pastoralists. To enter a grazing tract, the pastoralists had to show their pass and pay the tax.

→ How Did these Changes Affect the Lives of Pastoralists

• Wasteland Rules, Forest Acts, Criminal Tribes Act and imposition of grazing tax affected the lives of pastoralists adversely. These measures led to the serious shortage of pastures, as grazing lands were turned into cultivable land.
• The shepherds and cattle herders could no longer freely graze their cattle in the forests.
• Nomadic people had to move frequently from one place to another in search of pastures.
• Animal stock declined as under-fed cattle died in large numbers during droughts and famines.

→ How Did the Pastoralists Cope with these Changes?

• Pastoralists coped up with the changes in a variety of ways. Some pastoralists reduced the number of cattle in their herds, since there was not enough pasture to feed large numbers.
• Some pastoralists discovered new pastures when movement to old grazing grounds became difficult.
• Over the years, some richer pastoralists began buying land and settling down, giving up their nomadic life.
• Many poor pastoralists borrowed money from moneylenders to survive. Some of them became labourers, working on fields or in small towns. Yet, pastoralists not only continue to survive in many regions but their numbers have also expanded over recent decades.
• Many ecologists believe that in dry regions and in the mountains, pastoralism is still ecologically the most practical way of life.
• Such changes on pastoral communities were imposed not only in our country, but also all over the world.

→ Pastoralistn in Africa

• Africa is a continent where over half the world’s pastoral population lives.
• Even today, over 22 million Africans depend on some forms of pastoral activities for their livelihood.
• The different pastoral communities of Africa are Bedouins, Berbers, Maasai, Somali, Turkna and Boran.
• Most of them lived in semi-arid grasslands or arid deserts where rain fed agriculture is difficult.
• They raise cattle, camels, goats, sheep and donkeys and they sell milk, meat, animal skin and wool. Some of them earn through trade and transport.

→  Where have the Grazing Lands Gone?

• The Maasai are nomadic and pastoral people who depend on milk and meat for livelihood.
• The Maasai cattle herders live primarily in East Africa : 3,00,000 in Southern Kenya and another 1,50,000 in Tanzania.
• The new colonial laws and regulations took away their lives in times of drought and even reshaped their social relationships.
• One of the problems that Maasais have faced is the continuous loss <?f their grazing lands.
• In 1885, under colonial rule, Maasai land was cut into half with an International boundary between British Kenya and German Tonganyika. The Maasais lost about 60 per cent of their pre-colonial lands.
• As cultivation expanded, pasturelands were turned into cultivated fields.
• Large areas of grazing lands were also turned into game reserves like the Maasai Mara in Kenya and Serengeti Park in Tanzaniya; pastoralists were not allowed to enter these reserves.

→ The Borders are Closed

• Pastoral groups were forced to live within the boundaries of special reserves.
• They were not allowed to move out with their stock without special permits. They were not even allowed to trade and enter the markets in white areas.
• These restrictions adversely affected both their pastoral and trading activities. Earlier, pastoralists not only looked after animal herds, but they also traded in various products.

→ When Pastures Dry

• The Maasais were forced to live in drought-prone areas, as a result of this, large num-bers of Maasai cattle died of starvation and diseases. There was a severe decline in the animal stock of Maasais.

→ Not All were Equally Affected

• The colonial rules had unequal effects on elders and warrior groups of Maasai society.
• The British imposed various restrictions on raiding and warfare. Thus, the traditional authority of both elders and warriors was negatively affected.
• The chiefs appointed by the colonial government collected wealth overtime. They started to live in towns and got involved in trades. Their families stayed back in villages to look after land and animals.
• The poor pastoralists did not have the resources to survive in bad times and thus they were forced to do odd jobs like charcoal burners, workers in road and building construction etc.

→ Conclusion

• Pastoral communities in different parts of the world are affected in a variety of different way by changes in the modem world. New laws and new borders affect the patterns of their movement.
• They change the path of their annual movements reduce their cattle numbers, pressurise for rights to enter new areas.
• They put political pressure on the government for relief, subsidy and other forms of support and demand a right in the management of forests and water resources.
• Economists and environmentalists have increasingly come to recognise that pastoral nomadism is a form of life that is perfectly suited to many hilly and dry regions of the world even today.

→ Important Dates and Related Events

• 1871: The Criminal Tribes Act was passed by the colonial government in India.
• 1850s- 1880s: The right to collect the tax was auctioned out to contractors.
• 1880s: The government began collecting taxes directly from the pastoralists. Each of them was given a pass.
• 1885: The Maasailand (Africa) was cut into two with an international boundary between British Kenya and German Tanganyika.

→ Nomads: People who move from one place to another to earn their living.

→ Gujjar Bakarwals: A Nomadic tribe of Jammu and Kashmir.

→ Kafila: Movement of a large number of households on a long journey.

→ Gaddi: The shepherd community of Himachal Pradesh.

→ Bhabar: A dry forested area below the foothills of Garhwal and Kumaun.

→ Bugyal: Vast meadows in the high mountains.

→ Bhotiyas, Sherpas, Kinnauris: Pastoral communities of the Himalayan region.

→ Dhars : Pasture land on the high mountains.

→ Dhangars : Pastoral community of Maharashtra.

→ Kharif: The autumn crop, usually harvested between September and October.

→ Rabi: The spring crop, usually harvested after March.

→ Stubble: Lower ends of grain stalks left in the grounds after harvesting.

→ Gollas,Kurumas,Kurubas: Pastoral communities in Karnataka and Andhra Pradesh.

→ Raikas: A pastoral tribe in Rajasthan.

→ Banjaras: A group of graziers found in U.P., Punjab, Rajasthan, Madhya Pradesh and Maharashtra.

→ Maru: A local name for Rajasthan’s desert.

→ Geneologist: A person who recounts the history of the community by narrating the facts orally.

→ Pastures: Tracts of land that are covered with a variety of grasses, roots and herbs, and wild flowers.

→ Pastoralists: A group of people that earns its living by the rearing of animals like goats, sheep, buffaloes, cows and camels.

→ Customary rights : Rights that people are used to by custom and tradition.

→ Bedouins, Berbers, Maasai, Somali, Boran and Turkang: Pastoral communities of Africa.

JAC Class 9 Social Science Notes

Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.5 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.5

Question 1.
Sides of triangles are given below, Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:
1. 7 cm, 24 cm, 25 cm
2. 3 cm, 8 cm, 6 cm
3. 50 cm, 80 cm, 100 cm
4. 13 cm, 12 cm, 5 cm
Solution :
1. 7 cm, 24 cm, 25 cm.
Here, the longest side is 25 cm.
25² = 625 and 7² + 24² = 49 + 576 = 625
∴ 25² = 7² + 24²
Here, the square of the longest side equals the sum of squares of the other two sides. Hence, the triangle with sides 7 cm, 24 cm and 25 cm is a right triangle and the length of its hypotenuse, is 25 cm.

2. 3 cm, 8 cm, 6 cm
Here, the longest side is 8 cm.
8² = 64 and 3² + 6² = 9 + 36 = 45
∴ 8² ≠ 3² + 6²
Hence, the triangle with sides 3 cm, 8 cm and 6 cm is not a right triangle.

3. 50 cm, 80 cm, 100 cm
Here, the longest side is 100 cm.
100² = 10000 and
50² + 80² = 2500 + 6400 = 8900
∴ 100² ≠ 50² + 80²
Hence, the triangle with sides 50 cm, 80 cm and 100 cm is not a right triangle.

4. 13 cm, 12 cm, 5 cm
Here, the longest side is 13 cm.
13² = 169 and 12² + 5² = 144 + 25 = 169
∴ 13² = 12² + 5²
Here, the square of the longest side equals the sum of squares of the other two sides. Hence, the triangle with sides 13 cm. 12 cm and 5 cm is a right triangle and the length of its hypotenuse is 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM. MR.
Solution :

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR.
∴ ΔRMP ~ ΔPMQ ~ ΔRPQ (Theorem 6.7)
Now, ΔRMP ~ ΔPMQ
∴ \(\frac{PM}{QM}=\frac{RM}{PM}\)
∴ PM² = QM. MR

Question 3.
In the given figure, ABD is a triangle right-angled at A and AC ⊥ BD. Show that
1. AB² = BC.BD
2. AC² = BC.DC
3. AD² = BD.CD
Solution :

ABD is a triangle right angled at A and AC ⊥ BD.
∴ ΔBCA ~ ΔACD ~ ΔBAD (Theorem 6.7)
1. ΔBCA ~ ΔBAD
∴ \(\frac{AB}{DB}=\frac{CB}{AB}\)
∴ AB² = BC. BD

2. ΔBCA ~ ΔACD
∴ \(\frac{AC}{DC}=\frac{BC}{AC}\)
∴ AC² = BC. DC

3. ΔACD ~ ΔBAD
∴ \(\frac{AD}{BD}=\frac{CD}{AD}\)
∴ AD² = BD . CD

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution :
ABC is an isosceles triangle right angled at C.
Hence, AB is the hypotenuse and the other two sides are equal, i.e., BC = AC
In ΔABC, ∠C = 90°

∴ By Pythagoras theorem,
AB² = BC² + AC²
∴ AB² = AC² + AC² (∵ BC = AC)
∴ AB² = 2AC²

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.
Solution :
In ΔABC, AC = BC and AB² = 2AC²
AB² = 2AC²
∴ AB² = AC² + AC²
∴ AB² = AC² + BC² (∵ AC = BC)
Hence, by the converse of Pythagoras theorem, ΔABC is right triangle in which ∠C is a right angle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution :

In ΔABC, AB = BC = CA = 2a.
Let AD be its altitude
∴ ∠ADB = ∠ADC = 90°
In ΔADB and ΔADC,
∠ADB = ∠ADC = 90°
AB = AC
AD = AD
∴ By RHS criterion,
= ΔADC
∴ BD = CD
But, BD + CD = BC
∴ BD = CD = \(\frac{1}{2}\)BC = \(\frac{1}{2}\)(2a) = a
Now, in ΔADB, ∠D = 90°
∴ By Pythagoras theorem,
AB² = AD² + BD²
∴ (2a)² = AD² + (a)²
∴ 4a² – a² = AD²
∴ AD² = 3a²
∴ AD = \(\sqrt{3}\)a
All the altitudes of an equilateral triangle are equal.
Hence, each of the altitudes of equilateral ΔABC with side 2a is \(\sqrt{3}\)a.

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution :
Given: ABCD is a rhombus.
To prove : AB² + BC² + CD² + DA² = AC² + BD²

Proof: ABCD is a rhombus.
∴ AB = BC = CD = DA ……………(1)
Let its diagonals AC and BD intersect at M.
Then, MA = MC = \(\frac{1}{2}\)AC.
MB = MD = \(\frac{1}{2}\)BD and
∠AMB = ∠BMC = ∠CMD = ∠DMA = 90°
In ΔAMB, ∠AMB = 90°
∴ AB² = MA² + MB² (Pythagoras theorem)
∴ AB² = (\(\frac{AC}{2}\))² + (\(\frac{BD}{2}\))²
∴ AB² = (\(\frac{AC}{2}\))² + (\(\frac{BD}{2}\))²
∴ 4AB² = \(\frac{\mathrm{AC}^2}{4}+\frac{\mathrm{BD}^2}{4}\)
∴ 4AB² = AC² + BD²
∴ AB² + AB² + AB² + AB² = AC² + BD²
∴ AB² + BC² + CD² + DA² = AC² + BD²

Question 8.
In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:
1. OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
2. AF² + BD² + CE² = AE² + CD² + BF².
Solution :

Join OA, OB and OC.
Here, in ΔOFA and ΔOFB, ∠F = 90°, in ΔODB and ΔODC, ∠D = 90° and in ΔOEC and ΔOEA. ∠E = 90°.
Then, Pythagoras theorem is applicable in all the triangles.
1. In ΔOFA, ∠F = 90°
∴ OA² = OF² + AF²
∴ AF² = OA² – OF² …………..(1)
In ΔODB, ∠D = 90°
∴ OB² = OD² + BD²
∴ BD² = OB² – OD² …………..(2)
In ΔOEC, OE² + CE²
∴ CE² = OC² – OE² …………..(3)
Adding (1), (2) and (3).
AF² + BD² + CE² = OA² – OF² + OB² – OD² + OC² – OE²
∴ OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²

2. AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF²
∴ AF² + BD² + CE² = (OA² – OE²) + (OB² – OF²) + (OC² – OD²)
∴ AF² + BD² + CE² = AE² + BF² + CD² (∵ ΔOAE, ΔOBF and ΔOCD are right triangles)
∴ AF² + BD² + CE² = AE² + CD² + BF²

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution :

Here, AB is the wall with window at point A and AC is the ladder.
Then, AC = 10m and AB = 8 m.
In ΔABC, ∠B = 90°.
∴ AC² = AB² + BC² (Pythagoras theorem)
∴ 10² = 8² + BC²
∴ BC² = 10² – 8²
∴ BC² = 100 – 64
∴ BC² = 36
∴ BC = 6 m
Thus, the distance of the foot of the ladder from the base of the wall is 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution :

Here, AB is the vertical pole in which the guy wire is attached at point A and AC is the guy wire and 18 m the stake is attached to its end C.
Then, AC = 24 m and AB = 18 m.
In ΔABC, ∠B = 90°
∴ AC² = AB² + BC² (Pythagoras theorem)
∴ 24² = 18² + BC²
∴ BC² = 576 – 324
∴ BC² = 252
∴ BC² = 4 × 9 × 7
∴ BC = 2 × 3 × \(\sqrt{7}\)
∴ BC = 6\(\sqrt{7}\) m
Thus, the stake should be driven 6\(\sqrt{7}\)m far from the base of the pole, so as to make the wire taut.

Question 11.
An airplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another airplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1\(\frac{1}{2}\) hours?
Solution :

Here, A is the airport, B is the position of the first plane flying due north after 1\(\frac{1}{2}\) hours and C is the position of the second c- plane flying due west after 1\(\frac{1}{2}\) hours.
[Note: For the sake of simplicity, we consider that both the planes are flying at the same height and point A representing the airport is also imagined to be at the same height.]
Then, AB = distance covered by the first plane in 1\(\frac{1}{2}\) hours
= Speed × Time
= 1000 × \(\frac{3}{2}\)
= 1500 km
Similarly, AC = distance covered by the second plane in 1\(\frac{1}{2}\) hours
= Speed × Time
= 1200 × \(\frac{3}{2}\)
= 1800 km
Also, ∠BAC is the angle formed by north direction and west direction.
Hence ∠BAC = 90°
Now, in ΔABC, ∠A = 90°
∴ BC² = AB² + AC² (Pythagoras theorem)
∴ BC² = (1500)² + (1800)²
∴ BC² = 22500 + 32400
∴ BC² = 54900
∴ BC = \(\sqrt{100 \times 9 \times 61}\)
∴ BC = 300\(\sqrt{61}\) km
Thus, the two planes will be 300\(\sqrt{61}\) km apart from each other after 1\(\frac{1}{2}\) hours.

Question 12.
Two poles of heights 6m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution :

Here, AB and CD are two erect poles of height 6 m and 11 m respectively.
The distance between the feet of the poles is 12 m.
Then, AB = 6 m, BD = 12 m, CD = 11 m, ∠B = 90° and ∠D = 90°.
Draw AE || BC.
Then, in quadrilateral ABDE.
∠B = ∠D = ∠E = ∠A = 90°.
Hence, ABDE is a rectangle.
∴ ED = AB = 6m and AE = BD = 12 m.
Then, CE = CD – DE = 11 – 6 = 5m
Now, in ΔAEC, ∠E = 90°.
∴ AC² = AE² + CE² (Pythagoras theorem)
∴ AC² = 12² + 5²
∴ AC² = 144 + 25
∴ AC² = 169
∴ AC = 13 m
Thus, the distance between the tops of the vertical poles is 13 m.

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE²
Solution :
In ΔABC, ∠C is a right angle, point D lies on CA and point E lies on CB.

Then, all the four triangles BCD, BCA, ECD and ECA are right triangles and in each of them C is a right angle.
Hence, Pythagoras theorem is applicable in all the four triangles.
In ΔECA, AE² = EC² + CA² ……………..(1)
In ΔBCD, BD² = BC² + CD² ……………..(2)
In ΔBCA, AB² = BC² + CA² ……………..(3)
In ΔECD, DE² = EC² + CD² ……………..(4)
Adding (1) and (2).
AE² + BD² = EC² + CA² + BC² + CD²
= (BC² + CA²) + (EC² + CD²)
= AB² + DE² [By (3) and (4)]
Thus, AE² + BD² = AB² + DE²

Question 14.
The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3CD (see the given figure). Prove that 2AB² = 2AC² + BC².
Solution :

DB = 3CD
∴ BC = DB + CD = 3CD + CD
∴ BC = 4CD …………..(1)
In ΔADB, ∠D = 90°
∴ AB² = AD² + DB² (Pythagoras theorem) …………..(2)
In ΔADC, ∠D = 90°
∴ AC² = AD² + CD² (Pythagoras theorem) …………..(3)
Subtracting (3) from (2),
AB² – AC² = (AD² + DB²) – (AD² + CD²)
∴ AB² – AC² = DB² – CD²
∴ AB² – AC² = (DB + CD) (DB – CD)
∴ AB² – AC² = (BC) (3CD – CD)
∴ AB² – AC² = (BC) (2CD)
Multiplying the equation by 2, we get
2AB² – 2AC² = (BC) (4CD)
∴ 2AB² – 2AC² = (BC) (BC)
∴ 2AB² = 2AC² + BC² [By (1)]

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD² = 7AB².
Solution :
Given: In equilateral ΔABC, D is a point on BC such that BD = \(\frac{1}{3}\)BC.
To prove: 9AD² = 7AB²

Construction: Draw AM ⊥ BC, such that M lies on BC.
Proof: ΔABC is an equilateral triangle. Suppose, AB = BC = AC = a
In equilateral ΔABC, AM is an altitude.
∴ AM is a median.
∴ BM = \(\frac{1}{2}\)BC = \(\frac{1}{2}\)a
∴ BD = \(\frac{1}{3}\)BC. Hence, DC = \(\frac{2}{3}\)BC
BD = \(\frac{1}{3}\)BC = \(\frac{1}{3}\)a
DM = BM – BD = \(\frac{1}{2}\)a – \(\frac{1}{3}\)a = \(\frac{1}{6}\)a
In ΔAMB, ∠M = 90°
∴ AB² = AM² + BM²
∴ a² = AM² + \(\frac{1}{4}\)a²
∴ AM² = \(\frac{3}{4}\)a²
In ΔAMD, ∠M = 90°
∴ AD² = AM² + DM²

∴ 9AD² = 7a²
∴ 9AD2 = 7AB² (∵ AB = a)

Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution :

ABC is an equilateral triangle in which AD is an altitude.
Let AB = BC CA- a units.
In an equilateral triangle, an altitude is a median also.
∴ AD is a median.
∴ BD = \(\frac{1}{2}\)BC = \(\frac{a}{2}\)units
In ΔADB, ∠D = 90°
∴ AB² = AD² + BD²
∴ (a)² = AD² + (\(\frac{a}{2}\))²
∴ a² = AD² + \(\frac{a^2}{4}\)
∴ \(\frac{3}{4}\)a² = AD²
∴ 3a² = 4AD²
∴ 3 (side)² = 4 (altitude)²

Question 17.
Tick the correct answer and justify: In ΔABC, AB = 6\(\sqrt{3}\) cm, AC = 12 cm and BC= 6 cm. The angle B is:
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution :
In ΔABC, AB = 6\(\sqrt{3}\) cm = 10.38 cm (approx),
AC = 12 cm and BC = 6 cm
Here, AC is the longest side.
Then, 12² = 144 and
(6\(\sqrt{3}\))² + (6)² – 108 + 36 = 144
Thus, 12² = (6\(\sqrt{3}\))² + (6)²
Hence, by the converse of Pythagoras theorem, ΔABC is a right triangle in which the longest side AC is the hypotenuse and its opposite angle ∠B is a right angle.
Hence, the correct answer is (C) 90°.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.6 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.6

Question 1.
In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that \(\frac{QS}{SR}=\frac{PQ}{PR}\)
Solution :

Construction: Through Q, draw a line parallel to PS which intersects RP extended at M.
Proof: In ΔMQR, S and P are points on QR and MR respectively and PS || MQ.
∴ \(\frac{QS}{SR}=\frac{MP}{PR}\) (BPT) ……………(1)
Now, PS || MQ and PQ is their transversal.
∴ ∠SPQ = ∠PQM (Alternate angles) ……(2)
Similarly, PS || MQ and MR is their transversal.
∴ ∠RPS = ∠PMQ (Corresponding angles) …………….(3)
PS is the bisector of ∠QPR.
∴ ∠SPQ = ∠RPS …………….(4)
From (2), (3) and (4).
∠PQM = ∠PMQ.
∴ In ΔPMQ, MP = PQ ……….(5)
From (1) and (5), we get
\(\frac{QS}{SR}=\frac{PQ}{PR}\)

Question 2.
In the given figure, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that:
1. DM² = DN . MC
2. DN² = DM . AN
Solution :

In quadrilateral DMBN, ∠B = ∠M = ∠N = 90°.
Hence, DMBN is a rectangle.
∴ DN = MB ………….. (1)
and DM = NB ………….. (2)
Now, BD ⊥ AC
∴ ∠BDC = 90° and ΔBDC is a right triangle in which DM is altitude on hypotenuse BC.
Then, ΔBMD ~ ΔDMC ~ ΔBDC. (Theorem 6.7)
∴ \(\frac{DM}{CM}=\frac{BM}{DM}\)
∴ DM² = BM. CM
∴ DM² = DN. MC [By (1), DN = MB] [Result (1)]
Similarly, ΔADB is a right triangle in which DN is altitude on hypotenuse AB.
∴ ΔAND ~ ΔDNB ~ ΔADB (Theorem 6.7)
∴ \(\frac{DN}{BN}=\frac{AN}{DN}\)
∴ DN2 = BN. AN
∴ DN² = DM . AN [By (2), DM = NB] [Result (2)]

Question 3.
In the given figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC . BD.
Solution :

In ΔADC, ∠D = 90°
∴ AC² = AD² + DC²
∴ AD² = AC² – DC² ……………(1)
In ΔADB, ∠D = 90°
∴ AB² = AD² + DB²
∴ AD² = AB² – DB² ……………(2)
From (1) and (2).
AC² – DC² = AB² – DB²
∴ AC² = AB² + DC² – DB²
∴ AC² = AB2 + (BC + DB)² – DB²
∴ AC² = AB² + BC² + 2BC . DB + DB² – DB²
∴ AC² = AB² + BC² + 2BC . BD

Question 4.
In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² – 2BC . BD.
Solution :

In ΔADC, ∠D = 90°
∴ AC² = AD² + CD²
∴ AD² = AC² – CD² ……………….(1)
In ΔADB, ∠D = 90°
∴ AB² = AD² + BD²
∴ AD² = AB² – BD² ……………….(2)
From (1) and (2),
AC² – CD² = AB² -BD²
∴ AC² = AB² + CD² – BD²
∴ AC² = AB² + (BC – BD)² – BD²
∴ AC² = AB² + BC² – 2BC . BD + BD² – BD²
∴ AC² = AB² + BC² – 2BC . BD

Question 5.
In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
1. AC² = AD² + BC. DM + (\(\frac{BC}{2}\))²
2. AB² = AD² – BC. DM + (\(\frac{BC}{2}\))²
3. AC² + AB² = 2AD² + \(\frac{1}{2}\)BC²
Solution :

Here, ΔAMD, ΔAMC and ΔAMB are right triangles.
Also, since AD is a median, D is the midpoint of BC.
∴ CD = BD = \(\frac{BC}{2}\)
Moreover, DM = CM – CD and DM = BD – BM
1. In ΔAMC, ∠M = 90°
∴ AC² = AM² + CM²
∴ AC² = AM² + (DM + CD)²
∴ AC² = AM² + DM² + 2. DM.CD + CD²
∴ AC² = (AM² + DM²) + (2CD) (DM) + CD²
∴ AC² = AD² + BC.DM + (\(\frac{BC}{2}\))²
(∵ In ΔAMD, AD² = AM² + DM²)

2. In ΔAMB, ∠M = 90°
∴ AB² = AM² + BM²
∴ AB² = AM² + (BD – DM)²
∴ AB² = AM² + BD² – 2 BD.DM + DM²
∴ AB² = (AM² + DM²) – (2BD) · (DM) + BD²
∴ AB² = AD² – BC.DM + (\(\frac{BC}{2}\))²
(∵ In ΔAMD, AD² = AM² + DM²)

3. Now, adding the results of part (1) and (2)
AC² + AB² = AD² + BC. DM + (\(\frac{BC}{2}\))² + AD² – BC . DM + (\(\frac{BC}{2}\))²
∴ AC² + AB² = 2AD² + 2(\(\frac{\mathrm{BC}^2}{4}\))
∴ AC² + AB² = 2AD² + \(\frac{1}{2}\)BC²
[Note: In this result, if we replace BC by 2BD, we get the famous result know as Apollonius theorem : If AD is a median of ΔABC, then AB² + AC² = 2 (AD² + BD²).]

Question 6.
Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
Solution :
First of all, we prove Apollonius Theorem.
In ΔABC, let AD be a median and AM be an altitude as shown in the figure.
Then, AB² + AC²
= AM² + BM² + AM² + CM²
= 2AM² + (BD – MD)² + (CD + MD)²
= 2AM² + (BD – MD)² + (BD + MD)² (∵ CD = BD)
= 2AM² + 2BD² + 2MD²
= 2(AM² + MD²) + 2BD²
= 2AD² + 2BD²
∴ AB² + AC² = 2(AD² + BD²)

Let PQRS be a parallelogram in which the diagonals bisect each other at O.
Then, PO = RO = \(\frac{1}{2}\)PR and
QO = SO = \(\frac{1}{2}\)QS.
Now, in ΔPQR, QO is a median.
∴ PQ² + QR² = 2(QO² + PO²)
∴ PQ² + QR² = 2{(\(\frac{QS}{2}\))² + (\(\frac{PR}{2}\))²}
∴ PQ² + QR² = \(\frac{1}{2}\)(QS² + PR² ) ………….(1)
Similarly.
in ΔQRS, QR² + RS² = \(\frac{1}{2}\)(QR² + PR²) ……(2)
in ΔRSP, RS² + SP² = \(\frac{1}{2}\)(QS² + PR²) ……….(3)
in ΔSPQ, SP² + PQ² = \(\frac{1}{2}\)(QS² + PR²) ……….(4)
Adding (1), (2), (3) and (4), we get
2(PQ² + QR² + RS² + SP²) = 2(QS² + PR²)
∴ PQ² + QR² + RS² + SP² = QS² + PR²
Thus, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Question 7.
In the given figure, two chords AB and CD intersect each other at the point P. Prove that
1. ΔAPC ~ ΔDPB
2. AP . PB = CP . DP
Solution :

Here, ∠CAB = ∠CDB
(Angles in the same segment)
∴ ∠CAP = ∠BDP
Similarly,
∠ACD = ∠DBA
(Angles in the same segment)
∴ ∠ACP = ∠DBP
Now, in ΔAPC and ΔDPB,
∠CAP = ∠BDP and ∠ACP = ∠DBP.
∴ By AA criterion, ΔAPC ~ ΔDPB. (Result 1)
∴ \(\frac{AP}{DP}=\frac{CP}{BP}\)
∴ AP . PB = CP . DP (Result 2)

Question 8.
In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle.
Prove that
1. ΔPAC ~ ΔPDB
2. PA . PB = PC . PD
Solution :

In cyclic quadrilateral ACDB,
ΔACD + ∠ABD = 180°
Again, ∠ACD + ∠ACP = 180° (Linear pair)
∴ ∠ABD = ∠ACP
∴ ∠PBD = ∠PCA.
Similarly, ∠CAB + ∠CDB = 180° (Cyclic quadrilateral)
∠CAB + ∠CAP = 180° (Linear pair)
∴ ∠CDB = ∠CAP
∴ ∠PDB = ∠PAC
Now, in ΔPDB and ΔPAC,
∠PBD = ∠PCA
∠PDB = ∠PAC
∴ By AA criterion, ΔPAC ~ ΔPDB [Result (1)]
∴ \(\frac{PA}{PD}=\frac{PC}{PB}\)
∴ PA.PB = PC.PD [Result (2)]

Question 9.
In the given figure, D is a point on side BC of ΔABC such that \(\frac{BD}{CD}=\frac{AB}{AC}\). Prove that AD is the bisector of ∠BAC.
Solution :

Through B. draw a line parallel to AD to intersect CA extended at P.
Then, in ΔPBC, A and D are points on PC and BC respectively and PB || AD.
∴ \(\frac{PA}{AC}=\frac{BD}{CD}\)
Also, \(\frac{BD}{CD}=\frac{AB}{AC}\) (Given)
∴ PA = AB
Now, in ΔPAB, PA = AB
∴ ∠ABP = ∠APB …………..(1)
AD || BP and AB is their transversal.
∴ ∠ABP = ∠BAD (Alternate angles) …………..(2)
AD || BP and CP is their transversal.
∴ ∠APB = ∠CAD (Corresponding angles) …………..(3)
From (1), (2) and (3),
∠BAD = ∠CAD
Moreover, ∠BAD + ∠CAD = ∠BAC.
Hence, AD is the bisector of ∠BAC.

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away from her and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see the given figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Solution :

Here, ΔABC represents the initial position in which A is the tip of her fishing rod, C is the fly at the end of the string and B is the point directly under the tip of the rod.
Then, in ΔABC, ∠B = 90°, AB = 1.8m and BC = 2.4 m.
∴ AC² = AB² + BC² (Pythagoras theorem)
∴ AC² = (1.8)² + (2.4)²
∴ AC² = 3.24 + 5.76
∴ AC² = 9
∴ AC = 3 m
Hence, in the initial position, she has 3 m of string out.
Length of string pulled-in in 1 sec = 5 cm
∴ Length of string pulled-in in 12 sec = 60 cm = 0.6 m
Now, in the second position, the length of string AC = 3m-0.6 m = 2.4 m and AB = 1.8 m.
Again, AC² = AB² + BC²
∴ (2.4)² = (1.8)² + BC²
∴ BC² = (2.4)² – (1.8)²
∴ BC² = (2.4 + 1.8) (2.4 – 1.8)
∴ BC² = 4.2 × 0.6
∴ BC² = 2.52
∴ BC = \(\sqrt{2.52}\)
∴ BC= 1.59 m (approx)
Now, the horizontal distance of the fly from her
= BC + 1.2 m
= (1.59 + 1.2) m
= 2.79 m

JAC Board Class 9th Social Science Solutions History Chapter 1 The French Revolution

JAC Class 9th History The French Revolution InText Questions and Answers 

Activity (Page No. 5)

Question 1.
Explain why the artist has portrayed the nobleman as the spider and the peasant as the fly.
Answer:
The artist has depicted the nobleman as the spider and the peasant as the fly showing that flies work hard to get their food, they roam around here and there, whereas the spider creates a trap and catches the flies and receives food without any hard work.

In the same way, the nobleman had developed such a monarchy in the French society, in which the peasants used to arrange for livelihood with hard work and the nobleman used to get it as a tax without any hard work. Such depiction by the artist represents the exploiting class and the legal system created by him in the erstwhile French society.

Activity (Page No. 6)

Question 1.
Fill in the blank boxes in figure 4 with appropriate terms from among the following: Food riots, scarcity of grain, increased number of deaths, rising food prices, weaker bodies.
Answer:

Activity (Page No. 7)

Question 1.
What message is Young trying to convey here? Whom does he mean when he speaks of ‘slaves’? Who is he criticising? What dangers does he sense in the situation of 1787?
Answer:

1. Here Young is trying to convey the message that any social system based on in justice will not last long and its consequences can be disastrous for the exploiters.
2. The ‘Slaves’ mentioned by him are the peasants, servants and landless labourers who were the underprivileged and deprived sections of French society at that time.
3. He is criticising the complete social system and especially the noblemen and the clergy.
4. The danger sensed by him is violence from the underprivileged section on the noblemen, clergy and their families.

Activity (Page No. 8)

Question 1.
Representatives of the Third Estate take the oath raising their arms in the direction of Bailly, the President of the Assembly, standing on a table in the centre. Do you think that during the actual event Bailly would have stood with his back to the assembled deputies? What could have been David’s intention in placing Bailly (fig.5) the way he has done?
Answer:

1. No, I think that during the actual event, Bailly would not have stood with his back to the assembled deputies.
2. Placing Bailly in this way, David has showed his intention to support the assembled deputies towards Bailly.

Activity (Page No. 13)

Question 1.
Identify the symbols in Box 1 which stand for liberty, equality and fraternity.
Answer:
Liberty: The broken chain, Red Phrygian cap.
Equality: The winged woman, The law tablet.
Fraternity: The bundle of rods or fasces, National colours of France such as Blue-White-Red.

Question 2.
Explain the meaning of the painting of the Declaration of Rights of Man and Citizen, (Fig. 8) by reading only the symbols.
Answer:
in Fig. 8 a woman is shown wearing a blue, white and red dress. These colours are the national colours of France. The lady carrying the broken chain symbolises the freedom of the French citizens. On the other side, a winged woman is also shown which is a symbol of personification of the law. It expresses that all French citizens are law-abiding. Both pictures of women are shown on the Law Tablet which conveys the message that all are equal before the law.

Question 3.
Compare the political rights which the Constitution of 1791 gave to the citizens with Articles 1 and 6 of the Declaration (Source C). Are the two documents consistent? Do the two documents convey the same idea?
Answer:
The two documents are consistent and convey the same idea that human beings are born equal and all citizens are equal in front of the law. However, the constitution of 1791 did not give practical form to these ideas.

Question 4.
Which groups of French society would have gained from the Constitution of 1791? Which groups would have had reason to be dissatisfied? What developments does Marat (Source B) anticipate in the future?
Answer:

1. The Third Estate of French society would have gained the maximum benefit from the Constitution of 1791.
2. The First and Second Estates or the Aristocratic group would have the reason to be dissatisfied. They were forced to give up their privileges and a common tax was levied in return of all services.
3. Marat anticipates another revolution in which the poor will rebel against the rich persons of the third estate and overthrow them, just like they had done to the noblemen and clergy.

Question 5.
Imagine the impact of the events in France on neighbouring countries such as Prussia, Austria-Hungary or Spain, all of which were absolute monarchies. How would the kings, traders, peasants, nobles or members of the clergy here have reacted to the news of what was happening in France?
Answer:
The shadow of the events of France must also have fallen on the absolute monarchy countries of the neighbourhood such as Prussia, Austria-Hungry or Spain etc. These autocratic monarchy countries may have reduced the exploitation of citizens by reducing their autocracy. Because they may have feared that the incident in France may be repeated here as well.

The kings, noble, clergy and other privileged sections of these countries would become fearful that what has happened in France can happen in their country also. The peasant’s traders would welcome the developments in France and sympathise with the peasants and underprivileged sections of that country.

Activity (Page No. 15)

Question 1.
Look carefully at the painting and identify the objects which are political symbols you saw in Box 1 (broken chain, red cap, fasces, Charter of the Declaration of Rights). The Pyramid stands for equality, often represented by a triangle. Use the symbols to interpret the painting. Describe your impressions of the female figure of liberty.
Answer:

1. A red cap and the Charter of the Declaration of Rights in the lady’s hand are the political symbols.
2. The triangle shape of pyramid stands for equality because its three equal sides represent the three Estates of the French society and signifies that their rights and powers are equal. She is holding in the other hand – The Torch of Freedom.
3. The female figure of liberty signifies the true idea of freedom that women are equal to men. So, they should enjoy the same basic rights.

Activity (Page No. 16)

Question 1.
Compare the views of Desmoulins and Robespierre. How does each one understand the use of state force? What does Robespierre mean by ‘the war of liberty against tyranny’? How does Desmoulins perceive liberty? Refer once more to Source C. What did the constitutional laws on the rights of individuals lay down? Discuss your views on the subject in class.
Answer:
1. Desmoulins was all for a humanitarian view of democracy. He argued that a true democracy means that all persons should enjoy equal rights; everyone should enjoy the fruits of liberty. He condemned the use of force even against those who opposed the views of a democratic government.

2. Robespierre was trying to find justification for his tyranny against all resistance, by posing to be the saviour of the ideals of the revolution and the republic. For him, terror is nothing but justice and is used to meet the most urgent needs of the fatherland.

3. Desmoulins perceive liberty as happiness, reason, equality and Justice. It is the declaration of rights.

4. The constitutional laws on the rights of individuals lay down the following rights:
(a) Right to freedom,
(b) Right to equality,
(c) Right to protection of property
(d) Right against exploitation, and
(e) Right to expression.

Activity (Page No. 18)

Question 1.
Describe the persons represented in Fig. 12 their actions, their postures, the objects they are carrying. Look carefully to see whether all of them come from the same social group. What symbols has the artist included in the image? What do they stand for? Do the actions of the women reflect traditional ideas of how women were expected to behave in public? What do you think: does the artist sympathise with the women’s activities or is he critical of them? Discuss your views in class.
Answer:

1. The persons represented in Fig 12 are women of Paris (France). Most of them came from the Third Estate. Their actions, their postures represent a rebellious violent mood.
2. The artist included the following symbols in the image :
   (a) Hoes, swords and pitchforks to indicate power and rebellion.
   (b) A balance scale on top to indicate equality among men and women.
   (c) A drum to indicate announcement.
3. No, the actions of the women do not reflect traditional ideas of how women were expected to behave in public.
4. Yes, the artist sympathises with the women.

Activity (Page No. 20)

Question 1.
Compare the manifesto drafted by Olympe de Gouges (Source F) with the Declaration of the Rights of Man and Citizen (Source C).
Answer:
The Declaration of the Rights of Men and citizens declares only the rights of Men and Citizens. Rights of women are not discussed in any article of this Declaration. On the other hand, the Declaration prepared by Olympe de Gouges discusses the rights of both men and women on the basis of equality.

Question 2.
Imagine yourself to be one of the women in Fig. 13. Formulate a response to the arguments put forward by Chaumette (Source G.)
Answer:
The arguments given by Chaumette are not justifiable. He has mentioned only the biological role of women. As individual, women have equal rights to men and so they must be treated equal to men. They would of course, continue their traditional roles like feeding and nurturing children, things of the household etc. but women are equally capable as men to carry out the other duties of human society also.

Activity (Page No. 21)

Question 1.
Record your impression of this print (Fig. 14). Describe the objects lying on the ground. What do they symbolise? What attitude does the picture express towards non-European slaves?
Answer:

1. This print shows the bad condition of slaves during seventeenth century. The tri-colour banner on the top carries the slogan“The rights of man” and an inscription below says“The freedom of the unfree.”
2. The symbols of power sword, gun etc. are lying on the ground. It shows that slavery has been abolished.
3. The picture expresses the racial behaviour of a French woman towards non-European slaves. She is giving European clothes to them as a mark of superior culture.

Activity (Page No. 22)

Question 1.
Describe the picture in your own words. What are the images that the artist has used to communicate the following ideas: greed, equality, justice, takeover by the state of the assets of the church?
Answer:

1. Greed: Protested by a fat man.
2. Equality: Man and woman standing together.
3. Justice: Two persons are going with off mood, which signifies that they did not get proper justice.
4. Assets of the church taken over by the state. A man is shown pressed in between the force machine which signifies the confiscation of assets of the church by the state.

Activity (Page No. 24)

Question 1.
Find out more about anyone. of the revolutionary figures, you have read about in this chapter. Write a short biography of this person.
Answer:
Olympe de Gouges (1748-1793): Olympe de Gouges was one of the most important of politically active women in revolutionary France. She protested against the Constitution and the Declaration of Rights of Man and Citizen as they excluded women from basic rights that each human being was entitled to.

So, in 1791, she wrote a Declaration of Rights of Woman and Citizen, which she addressed to the Queen and to the members of the National Assembly, demanding that they act upon it. In 1793, Olympe de Gouges criticised the Jacobin government for forcibly closing down women’s clubs. She was tried by the National Convention, which charged her with treason. Soon after this, she was executed.

Question 2.
The French Revolution saw the rise of newspapers describing the events of each day and week. Collect information and pictures on any one event and write a newspaper article. You could also conduct an imaginary interview with important personages such as Mirabeau, Olympe de Gouges or Robespierre.
Answer:
An article in a newspaper at the end of Bastille: On the morning of 14th July 1789, the city of Paris was in a state of alarm. Rumours spread that king would soon order the army to open fire upon the  citizens. Some 7,000 men and women gathered in front of the town hall and decided to form a people’s militia. Finally, a group of several hundred people marched towards the fortress prison, the Bastille, where they hoped to find hoarded ammunition.

In the armed fight that followed, the Commander of the Bastille was killed and the prisoners were released, though there were only seven of them. The fortress was demolished and its stone fragments were sold in the market. The Bastille, which stood for the despotic rule of the king came to an end cowardly. It masked the end of feudalism and brought in a new era.

1. Interview with Mirabeau :
Journalist: Sir, you belong to the nobility, but here you are supporting those people who are against the nobility or aristocracy.

Mirabeau: Yes, I was born in an aristocratic group but it doesn’t mean that others are doing wrong and should not protest.

Journalist: Sir, will you support certain privileges based on birth which are given to some special classes of society?

Mirabeau: No, I am not convinced of the need to do with a society of feudal privileges by birth.

2. Interview with Olympe de Gouges :
Journalist: Madam, why do you oppose the Declaration of Rights of Man and Citizen?

Gouges: I am opposing this Declaration because there is no discussion of fundamental rights of women.

Journalist: What is the reason behind your criticism for the Jacobin government?

Gouges: I am criticising the Jacobin government for forcibly closing down the women’s clubs,

3. Interview Journalist Robespierre
Journalist: Sir, How would you establish and consolidate democracy?

Robespierre: To establish and consolidate democracy, I would annihilate the enemies of the republic at home and abroad.

Journalist: What method should be adopted by a democratic government during the revolution?

Robespierre: In the time of revolution, a democratic government may relay on terror.

Journalist: Sir, what do you mean by ‘Terror’?

Robespierre: Terror is nothing but justice. It is a swift, severe and inflexible policy which is used to meet the most urgent need of the fatherland.

JAC Class 9th History The French Revolution Textbook Questions and Answers 

Question 1.
Describe the circumstances leading to the outbreak of revolutionary protest in France.
Answer:
The following circumstances led to the outbreak of revolutionary protest in France:
1. Political Causes:
Emperor Louis XVI and his queen, Marie Antoinette, wasted’ money on their luxurious living and wasteful festivities. The high posts were often sold. The whole administration was corrupt and each department had its own laws. In the absence of any uniform system, there was confusion all around. The people were tired of such a rotten system of administration and wanted a change.

2. Social Causes :
The social conditions of France were as distressing as its political organisation. The clergy and the nobility led a luxurious life and enjoyed many privileges by birth, while the peasants and the labourers led a very hard life. They had to pay heavy taxes.

In addition to the direct taxes paid to the state, peasants had to pay taxes to the first and second estates also. They had often to be content with meals of boiled grass. It was an unfair situation which led to the growth of a feeling of discontent among the members of the Third Estate.

3. Econoir causes :
Long years of war had drained the financial resources of France, added to this was the cost of maintaining the extravagant court. To meet tht e expenses, the state was forced to increase taxes imposed on common people, At this time, there was a greater demand for food grains due to huge growth in population.

As a result, the price of bread rose. Due to rising prices and inadequate wages, most of the population could not even afford the basic means of livelihood. This led to a crisis of subsistence and an increase in the gap between the rich and the poor.

4. Influence of the Philosophers and Writers :
The middle class emerged educated and wealthy during the eighteenth century. The system of privileges as promoted by the feudal society was against their interest. Being educated, the members of this class had access to various ideas of equality and freedom proposed by the French and English political and social philosophers.

These ideas got popularised among the masses as a result of intensive discussions and debates in salons/halls and coffee-houses, and through books and newspapers. The ideas of philosophers played a great role. John Locke, Jean Jacques Rousseau and Montesquieu rejected the doctrine of divine kingship.

5. Immediate Cause:
On 5th May 1789, Louis XVI called together an assembly of the Estates General to pass proposals for new taxes. Voting in the Estates General in the past had been conducted according to the principle that each Estate had one vote. But this time, members of the third Estate demanded that voting now should be conducted by the assembly as a whole, where each member would have one vote. When the king rejected this proposal, members of the third Estate walked out of the assembly in protest, which was the spark of outbreak of the French revolution.

Question 2.
Which groups of French society benefitted from the revolution? Which groups were forced to relinquish power? Which sections of society would have been disappointed with the outcome of the revolution?
Answer:

1. All the groups of Third Estate were benefitted from the revolution. These groups included peasants, artisans, petty officers, lawyers, teachers, servants, merchants, doctors and traders.
2. The clergy and the nobility were forced to surrender power and all the privileges were taken away from them.
3. Feudal lords, nobles, clergy and women would have been disappointed with the outcome of the revolution.

Question 3.
Describe the legacy of the French Revolution for the people of the world during the nineteenth and twentieth centuries.
Answer:
The legacy of French Revolution for the people of the world during the nineteenth and twentieth centuries can be described as given below :

1. The French Revolution put an end to the arbitrary rule and developed the idea of Republic in Europe and in other parts of the world.
2. It was the first national movement that adopted the ideals of “liberty, equality and fraternity”. These ideas became the basic doctrine of democracy for every nation in the 19th and 20th century.
3. It marked the end of feudalism and brought in a new era of industrial capitalism.
4. It initiated social and political changes that took place in different parts of Europe.
5. It gave the term ‘Nation’ its modern meaning and promoted the concept of ‘nationalist’ which inspired the people in Poland, Germany, Netherlands and Italy to establish Nation-states in the countries.
6. The greatest effect was the starting of mass movements all over the world and the rise of a spirit of nationalism among the people.

Question 4.
Draw up a list of democratic rights we enjoy today whose origins could be traced to the French Revolution.
Answer:
Some of the democratic rights which we enjoy today whose origins could be traced to the French revolution are given in the Indian constitution. They are :
1. Right to Equality:
The Right to Equality has its origin in the French Revolution. Which was its guiding principle. In the Indian constitution, Right to Equality means equality before the law, prohibition of discrimination on the basis of religion, descent, caste, gender or place of birth and end of titles and equality of opportunity in matters of employment.

2. Right to Liberty or Freedom:
The origin of this right can also be traced to the French Revolution. In the premble to the Indian Constitution, Right to Liberty or Freedom means ‘Freedom of Speech, expression, education, protection of the life and physical freedom, belief, faith and worship.

3. Encouraging the Spirit of Fraternity:
The French Revolution introduced the growth of the spirit of fraternity and social welfare. In the Indian constitution, the concept of ‘Fraternity’ abolishes untouchability and local or provincial anti-social feelings.

4. Inspiring the Spirit of Democracy:
The French Revolution inspired the spirit of democracy which ensured many rights, viz., right against exploitation, right to life, right to vote etc., which we are enjoying today.

Question 5.
Would you agree with the view that the message of universal rights was beset with contradictions? Explain.
Answer:
Yes, I agree with the view that the message of universal rights, was beset with contradictions. Many ideals in the “Declaration of Right of Man and Citizen’ were not clear in their meanings. I can say this on the basis of the following examples:

1. “The law has the right to forbid only actions injurious to society”. It did not mention about criminal offences against individuals.
2. The declaration stated that, “Law is the expression of the general will. All citizens have the right to participate in its formation, personally or through their representatives. All citizens are equal before it.” Although, France became a constitutional monarchy, millions of citizens (men under the age of 25 including women) were still not allowed to vote at all.
3. Men who were above 25 years of age and who paid taxes equal to at least 3 days of a labourer’s wages, were given the status of active citizens, i.e., they had the right to vote. This was in striking contrast to the ideals that the revolution supported.
4. In the universal rights, nothing was mentioned about compulsory education for all.
5. It did not give the right to freedom of trade and occupation.

Question 6.
How would you explain the rise of Napoleon?
Answer:
The rise of Napoleon Bonaparte was an indirect result of the French Revolution. After the fall of the Jacobin government, a new constitution was introduced. It provided for two elected legislative councils, and a Directory, an executive made up of five members. However, the Directors often clashed with the legislative councils, who then sought to dismiss them. This clash was responsible for political instability. Napoleon took advantage of the situation and became a dictator with the help of the army.

In 1804, Napoleon declared himself the Emperor of France. He was a brilliant General and used his armies to conquer and dominate all the neighbouring countries except Britain and Russia. However, his successes were short-lived. Britain, Prussia, Austria and Russia jointly defeated him at Leipzig, and again at Waterloo in 1815. He was captured and sent as a prisoner to the island of St. Helena where he died in 1821.

JAC Class 9 Social Science Solutions

Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
1. 2x² – 7x + 3 = 0
2. 2x² + x – 4 = 0
3. 4x² + 4\(\sqrt{3}\)x + 3 = 0
4. 2x² + x + 4 = 0
Solution:
1. 2x² – 7x + 3 = 0
Dividing throughout by 2, we get

Thus, the roots of the given quadratic equation are 3 and \(\frac{1}{2}\).

2. 2x² + x – 4 = 0
Dividing throughout by 2, we get

Thus, the roots of the given quadratic equation are \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)

3. 4x² + 4\(\sqrt{2}\)x + 3 = 0
4x² + 4\(\sqrt{3}\)x + (\(\sqrt{3}\))² = 0
(2x + \(\sqrt{3}\))² = 0
(2x + \(\sqrt{3}\)) (2x + \(\sqrt{3}\)) = 0
2x + \(\sqrt{3}\) = 0 or 2x + \(\sqrt{3}\) =0
x = \(-\frac{\sqrt{3}}{2}\) or x = \(-\frac{\sqrt{3}}{2}\)
Thus, the roots of the given quadratic equation are \(-\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\)

4. 2x² + x + 4 = 0
Dividing throughout by 2, we get

But, the square of any real number cannot be negative.
Hence, the real roots of the given quadratic equation do not exist.

Question 2.
Find the roots of the quadratic equations given in Q.1 by applying the quadratic formula.
1. 2x² – 7x + 3 = 0
Here, a = 2, b = -7 and c = 3.
Then, b² – 4ac = (-7)² – 4(2)(3) = 25
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{7 \pm \sqrt{25}}{2(2)}\)
∴ x = \(\frac{7 \pm 5}{4}\)
∴ x = 3 or x = \(\frac{1}{2}\)
Thus, the roots of the given quadratic equation are 3 and \(\frac{1}{2}\).

2. 2x² + x – 4 = 0
Here, a = 2, b = 1 and c = -4.
Then, b² – 4ac = (1)² – 4(2)(-4) = 33
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-1 \pm \sqrt{33}}{2(2)}\)
∴ x = \(\frac{-1 \pm \sqrt{33}}{4}\)
Thus, the roots of the given quadratic equation are \(\frac{-1 \pm \sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)

3. 4x² + 4\(\sqrt{3}\)x + 3 = 0
Here, a = 4, b = 4\(\sqrt{3}\) and c = 3.
Then, b² – 4ac = (4\(\sqrt{3}\))² – 4(4)(3) = 0
Then, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-4 \sqrt{3} \pm \sqrt{0}}{2(4)}\)
∴ x = \(-\frac{\sqrt{3}}{2}\) or x = \(-\frac{\sqrt{3}}{2}\)
Thus, the roots of the given quadratic equation are \(-\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\).

4. 2x² + x + 4 = 0
Here, a = 2, b = 1 and c = 4.
Then, b² – 4ac = (1)² – 4 (2)(4) = -31 < 0
Since b² – 4ac < 0 for the given quadratic equation, the real roots of the given quadratic equation do not exits.

Question 3.
Find the roots of the following equations:
1. x – \(\frac{1}{x}\) = 3, x ≠ 0
2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\); x ≠ -4, 7
Solution:
1. x – \(\frac{1}{x}\) = 3, x ≠ 0
∴ x² – 1 = 3x
∴ x² – 3x – 1 = 0
Here, a = 1, b = -3 and c = -1.
Then, b² – 4ac = (-3)² – 4(1)(-1) = 13
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
x = \(\frac{3 \pm \sqrt{13}}{2(1)}\)
x = \(\frac{3 \pm \sqrt{13}}{2}\)
Thus, the roots of the given equation are \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\).

2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\); x ≠ -4, 7
∴ \(\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}\)
∴ \(\frac{-11}{x^2-3 x-28}=\frac{11}{30}\)
∴ \(\frac{-1}{x^2-3 x-28}=\frac{1}{30}\)
∴ -30 = x² – 3x – 28
∴ x² – 3x + 2 = 0
Here, a = 1, b = -3 and c = 2.
Then, b² – 4ac = (-3)² – 4(1)(2) = 1
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{3 \pm \sqrt{1}}{2(1)}\)
∴ x = \(\frac{3 \pm 1}{2}\)
∴ x = 2 or x = 1
Thus, the roots of the given equation are 2 and 1.
Note: Here, the method of factorisation would turn out to be more easier.

Question 4.
The sum of the reciprocals of Rehman’s ages (in years), 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:
Let the present age of Rehman be x years.
So, his age 3 years ago was (x – 3) years and his age 5 years hence will be (x + 5) years.
The sum of the reciprocals of these two ages (in years) is given to be \(\frac{1}{3}\).
∴ \(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)
∴ \(\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}\)
∴ 3(2x + 2) = (x – 3)(x + 5)
∴ 6x + 6 = x² + 2x – 15
∴ x² – 4x – 21 = 0
∴ x² – 7x + 3x – 21 = 0
∴ x(x – 7) + 3(x – 7) = 0
∴ (x – 7)(x + 3) = 0
∴ x – 7 = 0 or x + 3 = 0
∴ x = 7 or x = -3
Now, since x represents the present age of Rehman, it cannot be negative, i,e., x ≠ -3.
∴ x = 7
Thus, the present age of Rehman is 7 years.

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of those marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics be x.
Then, her marks in English = 30 – x, as her total marks in Mathematics and English is 30.
Had she scored 2 marks more in Mathematics and 3 marks less in English, her score in Mathematics would be x + 2 and in English would be 30 – x – 3 = 27 – x.
∴ (x + 2) (27 – x) = 210
∴ 27x – x² + 54 – 2x = 210
∴ -x² + 25x + 54 – 210 = 0
∴ -x² + 25x – 156 = 0
∴ x² – 25x + 156 = 0
∴ x² – 13x – 12x + 156 = 0
∴ x(x – 13) – 12(x – 13) = 0
∴ (x – 13)(x – 12) = 0
∴ x – 13 = 0 or x – 12 = 0
∴ x = 13 or x = 12
Then, 30 – x = 30 – 13 = 17 or
30 – x = 30 – 12 = 18
Thus, Shefall’s marks in Mathematics and in English are 13 and 17 respectively or 12 and 18 respectively.

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of the rectangular field be x m.
Then, it diagonal is (x + 60) m and the longer side is (x + 30) m.
In a rectangle, all the angles are right angles.
Hence, by Pythagoras theorem,
(Shorter side)² + (Longer side)² = (Diagonal)²
∴ x² + (x + 30)² = (x + 60)²
∴ x² + x² + 60x + 900 = x² + 120x + 3600
∴ x² – 60x – 2700 = 0
Here, a = 1, b = -60 and c = -2700.
Then, b² – 4ac = (-60)² – 4(1)(-2700)
= 3600 + 10800
= 14400
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{60 \pm \sqrt{14400}}{2(1)}\)
= \(\frac{60 \pm 120}{2}\)
∴ x = \(\frac{60+120}{2}\) or x = \(\frac{60-120}{2}\)
∴ x = 90 or x = -30
Since x denotes the shorter side of the rectangular field, x cannot be negative.
∴ x = 90
Then, x + 30 = 90 + 30 = 120
Thus, the shorter side (breadth) of the rectangular field is 90 m and the longer side (length) of the rectangular field is 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the smaller number be x.
Then, the larger number = \(\frac{x^2}{8}\)
Now, the difference of their squares is 180.
∴ \(\left(\frac{x^2}{8}\right)^2-(x)^2=180\)
∴ \(\frac{x^4}{64}-x^2=180\)
∴ x⁴ – 64x² – 11520 = 0
Let x² = y
∴ x⁴ = y²
∴ y² – 64y – 11520 = 0
Here, a = 1, b = -64 and c = -11520.
Then, b² – 4ac = (-64)² – 4(1)(-11520)
= 4096 + 46080
= 50176
Now, y = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{64 \pm \sqrt{50176}}{2(1)}\)
= \(\frac{64 \pm 224}{2}\)
∴ y = \(\frac{64+224}{2}\) or y = \(\frac{64-224}{2}\)
∴ y = 144 or y = -80
∴ x² = 144 or x² = -80
But, x² = -80 is not possible.
∴ x² = 144
∴ x = 12 or x = -12
Then, \(\frac{x^2}{8}=\frac{144}{8}=18\)
Thus, the required numbers are 12 and 18 or -12 and 18.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let the usual uniform speed of the train be x km/h.
∴ Time taken to travel 360 km at the speed of x km/h = \(\frac{\text { distance }}{\text { speed }}\) = \(\frac{360}{x}\) hours.
If the speed had been 5 km/h more, the new speed would be (x + 5) km/h and the time taken to travel 360 km at this increased speed would be \(\frac{360}{x+5}\) hours.
Now, New time = Usual time – 1
∴ \(\frac{360}{x+5}=\frac{360}{x}-1\)
∴ 360x = 360x + 1800 – x(x + 5) (Multiplying by x(x + 5))
∴ 0 = 1800 – x² – 5x
∴ x² + 5x – 1800 = 0
∴ x² + 45x – 40x – 1800 = 0
∴ x(x + 45) – 40(x + 45) = 0
∴ (x + 45) (x – 40) = 0
∴ x + 45 = 0 or x – 40 = 0
∴ x = -45 or x = 40
As, x is the speed (in km/h) of the train, x = -45 is not possible.
∴ x = 40
Thus, the usual uniform speed of the train is 40 km/h.

Question 9.
Two water taps together can fill a tank in \(9 \frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the time taken by the tap with smaller diameter to fill the tank be x hours.
Then, the time taken by the tap with larger diameter = (x – 10) hours.
So, the part of the tank filled in one hour by the tap with smaller diameter = \(\frac{1}{x}\) and by the tap with larger diameter = \(\left(\frac{1}{x-10}\right)\)
So, the part of tank filled in one hour by both the taps together = \(\frac{1}{x}+\frac{1}{x-10}\)
Both the taps together can fill the tank in \(9 \frac{3}{8}\) hours, i.e., \(\frac{75}{8}\) hours.
∴ The part of tank filled in one hour by both the tank together = \(\frac{8}{75}\)
∴ \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\).
∴ 75(x – 10) + 75x = 8x (x – 10) (Multiplying by 75x(x – 10))
∴ 75x – 750 + 75x = 8x² – 80x
∴ 8x² – 230x + 750 = 0
Here, a = 8, b = -230 and c = 750.
∴ b² – 4ac = (-230)² – 4 (8)(750)
= 52900 – 24000
= 28900
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{230 \pm \sqrt{28900}}{2(8)}\)
∴ x = \(\frac{230 \pm 170}{16}\)
∴ x = \(\frac{400}{16}\) or x = \(\frac{60}{16}\)
∴ x = 25 or x = 3.75
But, x ≠ 3.75, because for x = 3.75, x – 10 < 0.
∴ x = 25 and x – 10 = 15
Thus, the time taken by the tap with smaller diameter is 25 hours and that by the tap. with larger diameter is 15 hours.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of the passenger train be x km/h.
Then, the average speed of the express train is (x + 11) km/h.
∴ Time taken by passenger train to cover 132 km = \(\frac{132}{x}\) hours.
∴ Time taken by express train to cover 132 km = \(\frac{132}{x+11}\) hours.
Time taken by express train = Time taken by passenger train – 1
\(\frac{132}{x+11}=\frac{132}{x}-1\)
∴ 132x = 132(x + 11) – x(x + 11) (Multiplying by x(x + 11))
∴ 132x = 132x + 1452 – x² – 11x
∴ x² + 11x – 1452 = 0
Here, a = 1, b = 11 and c = -1452.
∴ b² – 4ac = (11)² – 4(1)(-1452)
= 121 + 5808 = 5929
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-11 \pm \sqrt{5929}}{2(1)}\)
∴ x = \(\frac{-11 \pm 77}{2}\)
x = 33 or x = -44
x = -44 is inadmissible as x represents the speed of the passenger train.
∴ x = 33 and x + 11 = 44
Thus, the speed of the passenger train is 33 km/h and the speed of the express train is 44 km/h.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the side of the smaller square be x m.
Then, the perimeter of the smaller square = 4x m
and the area of the smaller square = x² m².
From the given, the perimeter of the bigger square = (4x + 24) m.
∴ Side of the bigger square = \(\frac{4 x+24}{4}\) = (x + 6) m and hence, the area of the bigger square = (x + 6)² m².
∴ x² + (x + 6)² = 468
∴ x² + x² + 12x + 36 – 468 = 0
∴ 2x² + 12x – 432 = 0
∴ x² + 6x – 216 = 0
∴ x² + 18x – 12x – 216 = 0
∴ x(x + 18) – 12(x + 18) = 0
∴ (x + 18) (x – 12) = 0
∴ x + 18 = 0 or x – 12 = 0
∴ x = -18 or x = 12
Here, x = -18 is not possible as x represents the side of a square.
∴ x = 12 and x + 6 = 18
Thus, the side of the smaller square is 12 m and the side of the bigger square is 18 m.