Class 10

Jharkhand Board JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 12 Areas Related to Circles Exercise 12.3

Unless stated otherwise, use π = \(\frac{22}{7}\)

Question 1.
Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Solution:
Given, PQ = 24 cm, PR = 7 cm.
We know any triangle drawn from diameter RQ to the circle is 90°.
Here, ∠RPQ = 90°
In right ΔRPQ, RQ² = PR² + PQ² (By Pythagoras theorem)
RQ² = 7² + 24²
RQ² = 49 + 576
RQ² = 625
RQ = 25 cm
∴ Area of ΔRPQ = \(\frac{1}{2}\) × RP × PQ
= \(\frac{1}{2}\) × 7 × 24 = 84 cm²
∴ Area of semi-circle = \(\frac{\pi r^2}{2}=\frac{22}{7 \times 2}\left(\frac{25}{2}\right)^2\) (∵ r = \(\frac{\mathrm{PQ}}{2}=\frac{25}{2}\) cm)
= \(\frac{11 \times 625}{28}=\frac{6875}{28} \mathrm{~cm}^2\)
∴ Area of the shaded region = Area of the semi-circle – Area of right ΔRPQ
= \(\frac{6875}{28}-84\)
= \(\frac{6875-2352}{28}=\frac{4523}{28}\) cm².

Question 2.
Find the area of the shaded region in the figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:
R – radius of the bigger circle, r – radius of the smaller circle.
Area of the shaded portion = Area of sector OAC – Area of sector OBD

Question 3.
Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:
Area of the shaded portion = Area of the square – 2 × Area of one semicircle
= (14 × 14) – 2 × \(\frac{\pi r^2}{2}\) [∵ r = 7]
= 14 × 14 – 2 × \(\frac{22}{7} \times \frac{7 \times 7}{2}\)
= 196 – 154 = 42 cm².

Question 4.
Find the area of the shaded region in the figure, where a circular are of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Solution:
Area of the shaded portion = Area of the circle of radius 6 cm + Area of equilateral ΔABC – Area of the sector

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.

Solution:
Area of the shaded portion = Area of the square – Area of the 4 quadrants – Area of the circle
= (4 × 4) – 4 × area of one quadrant – area of the circle

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design.

Solution:
Area of the design (shaded region) = Area of the circle – Area of ΔABC
Area of equilateral triangle ABC = \(\frac{\sqrt{3} a^2}{4}\)
In ΔABC, AL ⊥ BC.

Alternative Method:

Area of shaded region
= 3[Area of segments]
= 3[Area of sector – Area of ΔOBC]
= 3[\(\pi r^2 \frac{\theta}{360^{\circ}}-\frac{1}{2}\) × BC × OL]

Question 7.
In the figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.

Solution:
Area of the shaded region = Area of the square – Area of the 4 quadrants
= Area of the square – 4 × area of one quadrant
= (14)² – 4 × \(\frac{1}{4}\)πr²
= (14)² – 4 × \(\frac{1}{4} \times \frac{22}{7}\) × 7 × 7
= 196 – 154
= 42 cm².

Question 8.
The figure depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m. long. If the track is 10 m wide, find:
(i) the distance around the track along its inner edge.
(ii) the area of the track.

Solution:
i) Distance around the inner track

ii) Area of the track:

Area of the track = l × b + l × b + 2\(\left[\frac{\pi \mathrm{R}^2}{2}-\frac{\pi \mathrm{r}^2}{2}\right]\) r = 30 mts, R = (30 + 10) = 40 mts.
= 106 × 10 + 106 × 10 + 2 × \(\frac{\pi}{2}\)(R² – r²)
= 1060 + 1060 + \(\frac{22}{7}\)[40² – 30²]
= 2120 + \(\frac{22}{7}\)[1600 – 900]
= 2120 + \(\frac{22}{7}\)[700]
= 2120 + 2200
= 4320 m².

Question 9.
In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
Given,
OA = 7 cm
∴ OD = 7 cm
Now, area of smaller circle whose diameter (OD) = 7 cm is

Now,
Area of ΔABC = \(\frac{1}{2}\) × AB × OC
= \(\frac{1}{2}\) × 2 × OA × OC
= \(\frac{1}{2}\) × 14 × 7 (∵ OA = OC)
= 49 cm².
Area of semi-circle ABCA = \(\frac{\pi r^2}{2}=\frac{22}{7 \times 2}(7)^2\)
= 77 cm²
∴ Area of segments BC and AC = Area of semi-circle – Area of AABC
= 77 – 49 = 28 cm²
∴ Area of total shaded region = Area of small circle + Area of segments BC and AC
= \(\frac{77}{2}\) + 28
= 38.5 + 28
= 66.5 cm².

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.73205).

Solution:
Area of the shaded portion = Area of equilateral ΔABC – Area of sector Axy – Area of sector Bxz – Area of sector Cyz.
π = 3.14, θ = 60°, r = ?, r = \(\frac{a}{2}\)
Area of the shaded portion = Area of the equilateral Δ – 3 × Area of one sector

Area of the shaded region = Area of ΔABC – \(\frac{3 \times \pi r^2 \theta}{360}\)
= 17320.5 – \(\frac{3 \times 3.14 \times 100 \times 100 \times 60}{360}\)
= 17320.5 – 1.57 × 10000
= 17320.5 – 15700.0
= 1620.5 sq.cm.

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.

Solution:
Number of circular designs = 9
Radius of the circular design = 7 cm
There are three circles in one side of the square handkerchief.
∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm
Area of the square = 42 × 42 cm² = 1764 cm²
Area of the circle = πr² = \(\frac{22}{7}\) × 7 × 7 = 154 cm²
Total area of the design = 9 × 154 = 1386 cm²
Area of the remaining portion of the handkerchief = Area of the square – Total area of the design
= 1764 – 1386
= 378 cm².

Question 12.
In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm find the area of the (i) quadrant OACB, (ii) shaded region.

Solution:

Question 13.
In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Solution:
Radius of the quadrant = Diagonal of the square (OB)
OB² = OA² + AB²
= 20² + 20²
= 400 + 400
= 800
OB2 = 400 × 2
OB = \(\sqrt{400 \times 2}\) = 20\(\sqrt{2}\)
Area of the shaded region = Area of the quadrant OPBQ – Area of the square OABC
= \(\frac{1}{4}\) πr² – (OA)²
= \(\frac{1}{4}\) × 3.14 × 20\(\sqrt{2}\) × 20\(\sqrt{2}\) – (20)²
= \(\frac{1}{4}\) × 3.14 × 400 × \(\sqrt{4}\) – 400
= \(\frac{1}{4}\) × 3.14 × 400 × 2 – 400
= 100 × 2 × 3.14 – 400
= 100 × 2(3.14 – 2)
= 200 × 1.14
= 228 sq.cm.

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ∠AOB = 30°, find the area of the shaded region.

Solution:
R = 21, r = 7, θ = 30°, π = \(\frac{22}{7}\)
Area of the shaded region = Area of sector OAB – Area of sector OCD

Question 15.
In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
BC² = 14² + 14²
BC² = 196 + 196 = 392
BC = \(\sqrt{392}\) = \(\sqrt{196 \times 2}\) = 14\(\sqrt{2}\) = d
r = \(\frac{14 \sqrt{2}}{2}=7 \sqrt{2}\)
Radius of the sector = 14.
Area of the shaded region Area of the semicircle BEC – Area of the segment BDCB

Question 16.
Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.

Solution:
Area of the design = Area of sector DXB + Area of ΔDCB
Area of the segment DXB = Area of the sector DXBC – Area of ΔDCB

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 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

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.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
1. x² – 3x – 10 = 0
2. 2x² + x – 6 = 0
3. \(\sqrt{2}\)x² + 7x + 5\(\sqrt{2}\) = 0
4. 2x² – x + \(\frac{1}{8}\) = 0
5. 100x² – 20x + 1 = 0
Solution:
1. x² – 3x – 10 = 0
∴ x² – 5x + 2x – 10 = 0
∴ x(x – 5) + 2(x – 5) = 0
∴ (x – 5)(x + 2) = 0
Hence, x – 5 = 0 or x + 2 = 0
∴ x = 5 or x = -2
Thus, the roots of the given equation are 5 and -2.
Verification:
For x = 5,
LHS = (5)² – 3(5) – 10
= 25 – 15 – 10
= 0
= RHS
For x = -2,
LHS = (2)² – 3(-2) – 10
= 4 + 6 – 10
= 0
= RHS
Hence, both the roots are verified.
Note that verification is not a part of the solution. It is meant only for your confirmation of receiving correct solution.

2. 2x² + x – 6 = 0
∴ 2x² + 4x – 3x – 6 = 0
∴ 2x (x + 2) – 3(x + 2) = 0
∴ (x + 2) (2x – 3) = 0
∴ x + 2 = 0 or 2x – 3 = 0
∴ x = -2 or x = \(\frac{3}{2}\)
Thus, the roots of the given equation are -2 and \(\frac{3}{2}\)

3. \(\sqrt{2}\)x² + 7x + 5\(\sqrt{2}\) = 0
∴ \(\sqrt{2}\)x² + 2x + 5x + 5\(\sqrt{2}\) = 0
∴ \(\sqrt{2}\)x(x + \(\sqrt{2}\)) + 5(x + \(\sqrt{2}\)) = 0
∴ (x + \(\sqrt{2}\))(\(\sqrt{2}\)x + 5) = 0
∴ x + \(\sqrt{2}\) = 0 or \(\sqrt{2}\)x + 5 = 0
∴ x = –\(\sqrt{2}\) or x = \(-\frac{5}{\sqrt{2}}\)
Thus, the roots of the given equation are –\(\sqrt{2}\) and \(-\frac{5}{\sqrt{2}}\)

4. 2x² – x + \(\frac{1}{8}\) = 0
∴ 16x² – 8x + 1 = 0 (Multiplying by 8)
∴ 16x² – 4x – 4x + 1 = 0
∴ 4x(4x – 1) -1 (4x – 1) = 0
∴ (4x – 1) (4x – 1) = 0
∴ 4x – 1 = 0 or 4x – 1 = 0
∴ x = \(\frac{1}{4}\) or x = \(\frac{1}{4}\)
Thus, the repeated roots of the given equation are \(\frac{1}{4}\) and \(\frac{1}{4}\)

5. 100x² – 20x + 1 = 0
100x² – 10x – 10x + 1 = 0
10x(10x – 1) -1 (10x – 1) = 0
(10x – 1)(10x – 1) = 0
10x – 1 = 0 or 10x – 1 = 0
x = \(\frac{1}{10}\) or x = \(\frac{1}{10}\)
Thus, the repeated roots of the given equation are \(\frac{1}{10}\) and \(\frac{1}{10}\).

Question 2.
Solve the problems given in Textual Examples:
1. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find out how many marbles they had to start with.
2. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. Find out the number of toys. produced on that day.
Solution:
1. Let the number of marbles that John had be x.
Then, the number of marbles that Jivanti had is (45 – x).
After losing 5 marbles, the number of marbles left with John = x – 5.
After losing 5 marbles, the number of marbles left with Jivanti = (45 – x) – 5 = 40 – x.
Therefore, the product of marbles with them is (x – 5) (40 – x), which is given to be 124.
Hence, we get the following equation:
(x – 5)(40 – x) = 124.
∴ 40x – x² – 200 + 5x = 124
∴ -x² + 45x – 324 = 0
∴ x² – 45x + 324 = 0
∴ x² – 36x – 9x + 324 = 0
∴ x(x – 36) – 9(x – 36) = 0
∴ (x – 36)(x – 9) = 0
∴ x – 36 = 0 or x – 9 = 0
∴ x = 36 or x = 9
Here, both the answers are admissible.
∴ 45 – x = 45 – 36 = 9 or
45 – x = 45 – 9 = 36
Thus, the number of marbles with John and Jivanti to start with are 36 and 9 respectively or 9 and 36 respectively.

2. Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy on that day = 55 – x.
So, the total cost of production (in rupees) on that day = x (55 – x).
Hence, x(55 – x) = 750
∴ 55x – x² – 750 = 0
∴ x² – 55x + 750 = 0
∴ x² – 30x – 25x + 750 = 0
∴ x(x – 30) – 25(x – 30) = 0
∴ (x – 30)(x – 25) = 0
∴ x – 30 = 0 or x – 25 = 0
∴ x = 30 or x = 25
Here, both the answers are admissible.
Hence, the number of toys produced on that day is 30 or 25.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the first of the two numbers whose sum is 27 be x.
Then, the second number is 27 – x and the product of those two numbers is x (27 – x).
Their product is given to be 182.
∴ x (27 – x) = 182
∴ 27x – x² – 182 = 0
∴ x² – 27x + 182 = 0
∴ x² – 14x – 13x + 182 = 0
∴ x(x – 14) – 13(x – 14) = 0
∴ (x – 14)(x – 13) = 0
∴ x – 14 = 0 or x – 13 = 0
∴ x = 14 or x = 13
Here, both the answers are admissible.
Hence, if x = 14, it gives that the first number = x = 14 and the second number = 27 – x = 27 – 14 = 13.
And if x = 13, it gives that the first number = x = 13 and the second number = 27 – x = 27 – 13 = 14.
Thus, in either case, the required numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let two consecutive positive integers be x and x + 1.
Then, the sum of their squares = (x)² + (x + 1)²
= x² + x² + 2x + 1
= 2x² + 2x + 1
This sum is given to be 365.
∴ 2x² + 2x + 1 = 365
∴ 2x² + 2x – 364 = 0
∴ x² + x – 182 = 0
∴ x² + 14x – 13x – 182 = 0
∴ x(x + 14) – 13(x + 14) = 0
∴ (x + 14)( x- 13) = 0
∴ x + 14 = 0 or x – 13 = 0
∴ x = -14 or x = 13
Since x is a positive integer, x = -14 is inadmissible.
∴ x = 13 and x + 1 = 13 + 1 = 14
Thus, the required consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of the right triangle be x cm.
Then, its altitude is (x – 7) cm.
The hypotenuse of the right triangle is given to be 13 cm.
Now, by Pythagoras theorem.
(Base)² + (Altitude)² = (Hypotenuse)²
∴ (x)² + (x – 7)² = (13)²
∴ x² + x² – 14x + 49 = 169
∴ 2x² – 14x – 120 = 0
∴ x² – 7x – 60 = 0
∴ x² – 12x + 5x – 60 = 0
∴ x(x – 12) + 5(x – 12) = 0
∴ (x – 12)(x + 5) = 0
∴ x – 12 = 0 or x + 5 = 0
∴ x = 12 or x = -5
As the base of a triangle cannot be negative. x = -5 is inadmissible.
Hence, x = 12.
Then, the base of the triangle = x = 12 cm and the altitude of the triangle = x – 7 = 12 – 7
= 5 cm.
Thus, the base and the altitude of the given triangle are 12 cm and 5 cm respectively.

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90. find the number of articles produced and the cost of each article.
Solution:
Let the number of pottery articles produced on that day be x.
Then, according to the given, the cost of production (in rupees) of each article = 2x + 3.
Hence, total cost of production (in rupees) on that day = x(2x + 3) = 2x² + 3x.
This total cost of production is given to be ₹ 90.
∴ 2x² + 3x = 90
∴ 2x² + 3x – 90 = 0
∴ 2x² – 12x + 15x – 90 = 0
∴ 2x(x – 6) + 15(x – 6) = 0
∴ (x – 6) (2x + 15) = 0
∴ x – 6 = 0 or 2x + 15 = 0
∴ x = 6 or x = –\(\frac{15}{2}\)
Here, x = –\(\frac{15}{2}\) is inadmissible as x represents the number of articles produced.
Hence, x = 6 and 2x + 3 = 2(6) + 3 = 15.
Thus, the number of pottery articles produced on that day is 6 and the cost of production of each article is ₹ 15.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.4

Question 1.
Which term of the AP 121, 117, 113, …. is its first negative term?
[Hint: Find n for a_n < 0]
Solution :
For the given AP 121, 117, 113, … a = 121 and d= 117 – 121 = – 4.
Let n^th term of the AP be its first negative term.
∴ a_n < 0
∴ a + (n – 1) d < 0
∴ 121 + (n – 1)(- 4) < 0
∴ 121 < 4 (n – 1) ∴ n > \(\frac{125}{4}\)
∴ n > 31\(\frac{1}{4}\)
Now, n being the number of a term is a positive integer and the smallest positive integer satisfying n > 31\(\frac{1}{4}\) is 32.
Hence, the 32nd term of the given AP is its first negative term.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Solution :
For the given AP, let the first term be a and the common difference be d.
a_n = a + (n – 1) d
∴ a₃ = a + 2d and a₇ = a + 6d
According to given information,
a₃ + a₇ = 6
∴ (a + 2d) + (a + 6d) = 6
∴ 2a + 8d = 6
∴ a + 4d = 3
∴ a = 3 – 4d …………….(1)
Again, the product of a₃ and a₇ is 8.
∴ (a + 2d) (a + 6d) = 8
∴ (3 – 4d + 2d) (3 – 4d + 6d) = 8 [by (1)]
∴ (3 – 2d) (3 + 2d) = 8
∴ 9 – 4d² = 8
∴ 1 = 4d²
∴ d² = \(\frac{1}{4}\)
∴ d = \(\frac{1}{2}\) or d = – \(\frac{1}{2}\)

Thus, the required sum of first sixteen terms is 76 or 20.

Question 3.
A ladder has rungs 25 cm apart. (see the given figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\)m apart, what is the length of the wood required for the rungs?
Solution :

The distance between the top rung and the bottom rung = 2\(\frac{1}{2}\)m = 250 cm.
The distance between two successive rungs = 25 cm.
∴ Total number of rungs = \(\frac{250}{25}\) + 1 = 11
Including the top rung as well as the bottom rung
The length of the first (bottom) rung = 45 cm.
The length of the 11th rung at top = 25 cm.
The length of rung decreases uniformly.
Hence, the lengths (in cm) of rungs form an AP in which the first term = 45 and 11th term = 25.
a_n = a+ (n – 1)d
∴ a₁₁ = a + 10 d
∴ 25 = 45 + 10 d
∴ – 20 = 10 d
∴ d = – 2
Thus, the length of rung uniformly decreases by 2 cm as we move from bottom to top.
Thus, the lengths (in cm) of rungs form a finite AP 45, 43, 41, ……….with 11 terms.
Then, sum of all the eleven terms will give the total length of the wood required for the rungs.
S_n = \(\frac{n}{2}\)(a + 1)
∴ S₁₁ = \(\frac{11}{2}\)(45 + 25)
∴ S₁₁ = \(\frac{11}{2}\) × 70
∴ S₁₁ = 385
Thus, the total length of the wood required for the rungs is 385 cm.

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: S_x-1 = S₄₉ – S_x]
Solution :
We know the sum of first n positive integers n(n+1) is given by S_n = \(\frac{n(n+1)}{2}\)
According to data, 1 + 2 + 3 + ……… + (x – 1) = (x + 1) + (x + 2) + ……….. + 49
∴ \(\frac{(x-1) \cdot x}{2}\) = (1 + 2 + 3 + … + 49) – (1 + 2 + 3 + … + x)
[Adding and subtracting (1 + 2 + 3 + … + x)}
∴ \(\frac{(x-1)(x)}{2}=\frac{49 \times 50}{2}-\frac{x(x+1)}{2}\)
∴ x(x – 1) + x(x + 1) = 49 × 50
∴ x² – x + x² + x = 49 × 50
∴ 2x² = 49 × 50
∴ x² = \(\frac{49 \times 50}{2}\)
∴ x² = 49 × 25
∴ x = 7 × 5
∴ x = 35
Thus, the value of x is 35.

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see the given figure). Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³]
Solution :

Volume of concrete required to build the first step = 50 × \(\frac{1}{2}\) × \(\frac{1}{4}\)m³ = \(\frac{25}{4}\)m³
Volume of concrete required to build the second step = 50 × \(\frac{1}{2}\) × (\(\frac{1}{4}\) + \(\frac{1}{4}\))m³ = \(\frac{25}{2}\)m³
Volume of concrete required to build the third step = 50 × \(\frac{1}{2}\) × (\(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\))m³ = = \(\frac{75}{4}\)m³ and so on up to 15 steps.
Thus, the volumes (in m³) of concrete required to build those 15 steps form the finite AP \(\frac{25}{4}\), \(\frac{25}{2}\), \(\frac{75}{4}\) …with 15 terms.
The sum of all the fifteen terms will give the quantity of total concrete required.
Here, a = \(\frac{25}{4}\), d = \(\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\) and n = 15.
S_n = \(\frac{n}{2}\)[(2a + (n – 1) d]
∴ S₁₅ = \(\frac{15}{2}\)[\(\frac{25}{2}\) +(15 – 1)\(\frac{25}{4}\)]
∴ S₁₅ = \(\frac{15}{2}\) [latex]\frac{25}{2}+\frac{175}{2}[/latex]
∴ S₁₅ = \(\frac{15}{2} \times \frac{200}{2}\)
∴ S₁₅ = 15 × 50
∴ S₁₅ = 750
Thus, 750 m³ of concrete is required to build the terrace.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
1. The taxi fare after each km, when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
2. The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
3. The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
4. The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.
Solution:
1. Here, the fare for 1 km = ₹ 15
the fare for 2 km = ₹ 15 + ₹ 8 = ₹ 23,
the fare for 3 km = ₹ 15 + 2 (₹8) = ₹ 31,
the fare for 4 km = ₹ 15 + 3 (₹8) = ₹ 39, and so on.
The list of numbers formed is 15, 23, 31, 39, …………..
Here, a₂ – a₁ = 23 – 15 = 8.
a₃ – a₂ = 31 – 23 = 8.
a₄ – a₃ = 39 – 31 = 8, and so on.
Thus, a_k+1 – a_k is the same every time.
Hence, the list of numbers forms an AP with a = 15 and d = 8.

2. Let the volume of air present in the cylinder at the beginning be V units. Then, volume of air remaining in the cylinder after first attempt = \(\frac{3}{4}\)V units. Also, volume of air remaining in the cylinder after second attempt = \(\left(\frac{3}{4}\right)^2\)V units. Here, the list of numbers formed is V, \(\frac{3}{4}\)V, \(\left(\frac{3}{4}\right)^2\)V, ………
Now, a₂ – a₁ = \(\frac{3}{4}\)V – V = –\(\frac{1}{4}\)V
a₃ – a₂ = \(\left(\frac{3}{4}\right)^2\)V – \(\frac{3}{4}\)V
= \(V\left(\frac{9}{16}-\frac{3}{4}\right)\)
= \(– \frac{3}{16}\)V
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the list of numbers does not form an AP.

3. Cost of digging first metre = ₹ 150
Cost of digging the second metre = ₹ 150 + ₹ 50
= ₹ 200
Cost of digging the third metre = ₹ 200 + ₹ 50
= ₹ 250
Cost of digging the fourth metre = ₹ 250 + ₹ 50
= ₹ 300
The list of numbers formed is 150, 200, 250, 300,…
Here, a₂ – a₁ = 200 – 150 = 50,
a₃ – a₂ = 250 – 200 = 50,
a₄ – a₃ = 300 – 250 = 50, and so on.
Thus, a_k+1 – a_k is the same every time. Hence, the list of numbers forms an AP with a = 150 and d = 50.

4. The formula of compound interest is known to us.
A = \(P\left(1+\frac{R}{100}\right)^T\)
Here, P = ₹ 10,000; R = 8% and T = 1, 2, 3, 4, ……….
Amount at the end of 1st year = ₹ 10000 (1.08).
Amount at the end of 2nd year = ₹ 10000 (1.08)².
Amount at the end of 3rd year = ₹ 10000 (1.08)³.
The list of numbers formed is 10000 (1.08), 10000 (1.08)2, 10000 (1.08), ………..
a₂ – a₁ = 10000 (1.08)² – 10000 (1.08)³
= 10000 (1.08) (1.08 – 1)
= 10000 (1.08) (0.08)
a₃ – a₂ = 10000 (1.08)³ – 10000 (1.08)²
= 10000 (1.08)² (1.08 – 1)
= 10000 (1.08)² (0.08)
Thus, a₂ – a₁ ≠ a₃ – a₂
Hence, the list of numbers does not form an AP.

Question 2.
Write first four terms of the AP when the first term a and the common difference d are given as follows:
1. a = 10, d = 10
2. a = -2, d = 0
3. a = 4, d = -3
4. a = -1, d = \(\frac{1}{2}\)
5. a = -1.25, d = -0.25
Solution:
1. a = 10, d = 10
First term a = 10
Second term = First term + d
= 10 + 10 = 20
Third term = Second term + d
= 20 + 10 = 30
Fourth term = Third term + d
= 30 + 10 = 40
Thus, the required first four terms of the AP are 10, 20, 30, 40.

2. a = -2, d = 0
First term = a = -2
Second term = First term + d
= -2 + 0 = -2
Third term = Second term + d
= -2 + 0 = -2
Fourth term = Third term + d
= -2 + 0 = -2
Thus, the required first four terms of the AP are -2, -2, -2, -2.

3. a = 4, d = -3
First term = a = 4
Second term = First term + d
= 4 + (-3) = 1
Third term = Second term + d
= 1 + (-3) = -2
Fourth term = Third term + d
= (-2) + (-3) = -5
Thus, the required first four terms of the AP are 4, 1, -2, -5.

4. a = -1, d = \(\frac{1}{2}\)
First term = a = -1
Second term = First term + d
= -1 + \(\frac{1}{2}\) = –\(\frac{1}{2}\)
Third term = Second term + d
= \(-\frac{1}{2}+\frac{1}{2}\) = 0
Fourth term = Third term + d
= 0 + \(\frac{1}{2}\) = \(\frac{1}{2}\)
Thus, the required first four terms of the AP are -1, – \(\frac{1}{2}\), 0, \(\frac{1}{2}\).

5. a = -1.25, d = -0.25
The general form of an AP is a, a + d, a + 2d, a + 3d, …….. Then,
First term = a = -1.25
Second term = a + d
= -1.25 + (-0.25) = -1.50
Third term = a + 2d
= -1.25 + 2(-0.25) = -1.75
Fourth term = a + 3d
= -1.25 + 3(-0.25) = -2.00
Thus, the required first four terms of the AP are -1.25, -1.50, -1.75, -2.00.

Question 3.
For the following APs, write the first term and the common difference:
1. 3, 1, 1, -3, ……..
2. -5, -1, 3, 7, …….
3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
4. 0.6, 1.7, 2.8, 3.9, …
Solution:
1. 3, 1, -1, -3,….
First term a = 3
Common difference d = a₂ – a₁ = 1 – 3 = -2

2. -5, 1, 3, 7,…..
First term a = -5
Common difference d = (-1) – (-5) = 4

3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
First term a = \(\frac{1}{3}\)
Common difference d = \(\frac{5}{3}-\frac{1}{3}=\frac{4}{3}\)

4. 0.6, 1.7, 2.8, 3.9, ……
First term a = 0.6
Common difference d = 1.7 – 0.6 = 1.1

Question 4.
Which of the following are APs? If they form an AP find the common difference d and write three more terms:
1. 2, 4, 8, 16, …….
2. 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …….
3. -1.2, -3.2, -5.2, -7.2, …….
4. -10, -6, -2, 2, …….
5. 3, 3 + 2\(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\), ….
6. 0.2, 0.22, 0.222, 0.2222, ……..
7. 0, -4, -8, -12, ……..
8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \cdots\)
9. 1, 3, 9, 27, ……
10. a, 2a, 3a, 4a, …….
11. a, a², a³, a⁴, …….
12. \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\), ……
13. \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\), …..
14. 1², 3², 5², 7², …….
15. 1², 5², 7², 7², …….
Solution:
2, 4, 8, 16, ….
a₂ – a₁ = 4 – 2 = 2,
a₃ – a₂ = 8 – 4 = 4,
a₄ – a₃ = 16 – 8 = 8
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

2. 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …
a₂ – a₁ = \(\frac{5}{2}-2=\frac{1}{2}\)
a₃ – a₂ = \(3-\frac{5}{2}=\frac{1}{2}\)
a₄ – a₃ = \(\frac{7}{2}-3=\frac{1}{2}\)
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = \(\frac{1}{2}\)
Next three terms are given by-
a₅ = \(\frac{7}{2}+\frac{1}{2}=4\),
a₆ = \(4+\frac{1}{2}=\frac{9}{2}\) and
a₇ = \(\frac{9}{2}+\frac{1}{2}=5\)

3. -1.2, -3.2, -5.2, -7.2, …..
a₂ – a₁ = -3.2 – (-1.2) = -2
a₃ – a₂ = -5.2 – (-3.2) = -2
a₄ – a₃ = -7.2 – (-5.2) = -2
Here, a_k+1 – a_k is the same everywhere. So, the given list of numbers forms an AP with d = -2.
Next three terms are given by-
a₅ = -7.2 + (-2) = -9.2.
a₆ = -9.2 + (-2) = -11.2 and
a₇ = -11.2 + (-2) = -13.2

4. -10, -6, -2, 2, ……..
a₂ – a₁ = (-6) – (-10) = 4.
a₃ – a₂ = (-2) – (-6) = 4.
a₄ – a₃ = 2 – (-2) = 4
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = 4.
Next three terms are given by-
a₅ = 2 + 4 = 6,
a₆ = 6 + 4 = 10 and
a₇ = 10 + 4 = 14

5. 3, 3+\(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\), …
a₂ – a₁ = 3 + \(\sqrt{2}\) – 3 = \(\sqrt{2}\),
a₃ – a₂ = (3 + 2\(\sqrt{2}\)) – (3 + \(\sqrt{2}\)) = \(\sqrt{2}\)
a₄ – a₃ = (3 + 3\(\sqrt{2}\)) – (3 + 2\(\sqrt{2}\)) = \(\sqrt{2}\)
Here, a_k+1 – a_k is the same everywhere.
So, the given list of numbers forms an AP with d = \(\sqrt{2}\).
Next three terms are given by-
a₅ = (3 + 3\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 4\(\sqrt{2}\),
a₆ = (3 + 4\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 5\(\sqrt{2}\) and
a₇ = (3 + 5\(\sqrt{2}\)) + \(\sqrt{2}\) = 3 + 6\(\sqrt{2}\)

6. 0.2, 0.22, 0.222. 0.2222,…
a₂ – a₁ = 0.22 – 0.2 = 0.02,
a₃ – a₂ = 0.222 – 0.22 = 0.002
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the given list of numbers does not form an AP.

7. 0, -4, -8, -12, ….
a₂ – a₁ = (-4) – 0 = -4,
a₃ – a₂ = (-8) – (-4) = -4,
a₄ – a₃ = (-12) – (-8) = -4
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = -4.
Next three terms are given by-
a₅ = (-12) + (-4) = -16,
a₆ = (-16) + (-4) = -20 and
a₇ = (-20) + (-4) = -24.

8.

Here, a_k+1 – a_k is the same everywhere. Hence, the given list of numbers forms an AP with d = 0.
Next three terms are given by-

9. 1, 3, 9, 27, ….
a₂ – a₁ = 3 – 1 = 2,
a₃ – a₂ = 9 – 3 = 6
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

10. a, 2a, 3a, 4a…
a₂ – a₁ = 2a – a = a,
a₃ – a₂ = 3a – 2a = a.
a₄ – a₃ = 4a – 3a = a
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of unknown numbers forms an AP with d = a.
Next three terms are given by-
a₅ = 4a + a = 5a,
a₆ = 5a + a = 6a and
a₇ = 6a + a = 7a

11. a, a², a³, a⁴,….
a₂ – a₁ = a² – a = a (a – 1),
a₃ – a₂ = a³ – a² = a² (a – 1)
Here, a₂ – a₁ ≠ a₃ – a₂.
Hence, the given list of unknown numbers does not form an AP.

12. \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\), ….
We know, \(\sqrt{8}\) = \(\sqrt{4 \times 2}\) = 2\(\sqrt{2}\).
\(\sqrt{18}\) = \(\sqrt{9 \times 2}\) = 3\(\sqrt{2}\) and
\(\sqrt{32}\) = \(\sqrt{16 \times 2}\) = 4\(\sqrt{2}\).
Hence, the given list of numbers is
\(\sqrt{2}\), 2\(\sqrt{2}\), 3\(\sqrt{2}\), 4\(\sqrt{2}\), ….
a₂ – a₁ = 2\(\sqrt{2}\) – \(\sqrt{2}\) = \(\sqrt{2}\),
a₃ – a₂= 3\(\sqrt{2}\) – 2\(\sqrt{2}\) = \(\sqrt{2}\)
a₄ – a₃ = 4\(\sqrt{2}\) – 3\(\sqrt{2}\) = \(\sqrt{2}\)
Here, a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = \(\sqrt{2}\).
Next three terms are given by-
a₅ = 4\(\sqrt{2}\) + \(\sqrt{2}\) = 5\(\sqrt{2}\) = \(\sqrt{50}\).
a₆ = 5\(\sqrt{2}\) + \(\sqrt{2}\) = 6\(\sqrt{2}\) = \(\sqrt{72}\) and
a₇ = 6\(\sqrt{2}\) + \(\sqrt{2}\) = 7\(\sqrt{2}\) = \(\sqrt{98}\).

13. \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\), ……
a₂ – a₁ = \(\sqrt{6}\) – \(\sqrt{3}\) = \(\sqrt{3}\)(\(\sqrt{2}\) – 1)
a₃ – a₂ = \(\sqrt{9}\) – \(\sqrt{6}\) = \(\sqrt{3}\)(\(\sqrt{3}-\sqrt{2}\))
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

14. 1², 3², 5², 7², ….
a₂ – a₁ = 3² – 1² = 9 – 1 = 8,
a₃ – a₂ = 5² – 3² = 25 – 9 = 16
Here, a₂ – a₁ ≠ a₃ – a₂
Hence, the given list of numbers does not form an AP.

15. 1², 5², 7², 7²,…
a₂ – a₁ = 5² – 1² = 25 – 1 = 24,
a₃ – a₂ = 7² – 5² = 49 – 25 = 24,
a₄ – a₃ = 73 – 7² = 73 – 49 = 24.
Here. a_k+1 – a_k is the same everywhere.
Hence, the given list of numbers forms an AP with d = 24.
Next three terms are given by-
a₅ = 73 + 24 = 97,
a₆ = 97 + 24 = 121 and
a₇ = 121 + 24 = 145.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.1

Question 1.
Check whether the following are quadratic equations:
1. (x + 1)² = 2(x – 3)
2. x² – 2x = (-2)(3 – x)
3. (x – 2)(x + 1) = (x – 1)(x + 3)
4. (x – 3)(2x + 1) = x(x + 5)
5. (2x – 1)(x – 3) = (x + 5)(x – 1)
6. x² + 3x + 1 = (x – 2)²
7. (x + 2) = 2x(x² – 1)
8. x³ – 4x² – x + 1 = (x – 2)³
Solution:
1. Here, LHS = (x + 1)² = x² + 2x + 1 and
RHS = 2(x – 3) = 2x – 6.
Hence, (x + 1)² = 2(x – 3) can be rewritten as x² + 2x + 1 = 2x – 6
∴ x² + 2x + 1 – 2x + 6 = 0
∴ x² + 7 = 0
It is of the form ax² + bx + c = 0.
(a = 1, b = 0, c = 7)
Hence, the given equation is a quadratic equation.

2. Here, RHS = (-2)(3 – x) = -6 + 2x.
Hence, x² – 2x = (-2) (3 – x) can be rewritten as
x² – 2x = -6 + 2x
∴ x² – 4x + 6 = 0
It is of the form ax² + bx + c = 0.
(a = 1, b = -4, c = 6)
Hence, the given equation is a quadratic equation.

3. Here, LHS = (x – 2) (x + 1) = x² – x – 2 and
RHS = (x – 1)(x + 3) = x² + 2x – 3.
Hence, (x – 2)(x + 1)(x – 1)(x + 3) be rewritten as
x² – x – 2 = x² + 2x – 3
∴ -3x + 1 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

4. Here, LHS = (x – 3) (2x + 1) = 2x² – 5x – 3
and RHS = x(x + 5) = x² + 5x.
Hence, (x – 3) (2x + 1) = x(x + 5) can be rewritten as
2x² – 5x – 3 = x² + 5x
∴ x² – 10x – 3 = 0
It is of the form ax² + bx + c = 0.
(a = 1, b = -10, c = -3)
Hence, the given equation is a quadratic equation.

5. Here, LHS = (2x – 1)(x – 3) = 2x² – 7x + 3 and
RHS = (x + 5) (x – 1) = x² + 4x – 5.
Hence, the given equation can be rewritten as
2x² – 7x + 3 = x² + 4x – 5
∴ x² – 11x + 8 = 0
It is of the form ax2 + bx + c = 0.
(a = 1, b = -11, c = 8)
Hence, the given equation is a quadratic equation.

6. Here, RHS = (x – 2)² = x² – 4x + 4
Hence, the given equation can be rewritten as
x² + 3x + 1 = x² – 4x + 4
∴ 7x – 3 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

7. Here, LHS = (x + 2)³ = x³ + 6x² + 12x + 8 and
RHS = 2x (x² – 1) = 2x³ – 2x.
Hence, the given equation can be rewritten as
x³ + 6x² + 12x + 8 = 2x³ – 2x
∴ -x³ + 6x² + 14x + 8 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

8. Here, RHS = (x – 2)³ = x³ – 6x² + 12x – 8.
Hence, the given equation can be rewritten as
x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
2x² – 13x + 9 = 0
It is of the form ax² + bx + c = 0.
(a = 2, b = -13, c = 9)
Hence, the given equation is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations:
1. The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Let the breadth (in metres) of the rectangular plot be x.
Then, the length (in metres) of the rectangular plot is 2x + 1.
Area of the rectangular plot = Length × Breadth
∴ 528 = (2x + 1) × x
(∵ Area is given to be 528 m²)
∴ 528 = 2x² + x
∴ 2x² + x – 528 = 0 is the required quadratic equation to find the length (2x + 1 m) and breadth (x m) of the rectangular plot.

2. The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let the two consecutive positive integer be x and x + 1.
Then, their product = x(x + 1) = x² + x.
This product is given to be 306.
∴ x² + x = 306
∴ x² + x – 306 = 0 is the required quadratic equation to find the consecutive positive integers x and x + 1.

3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Solution:
Let Rohan’s present age (in years) be x.
Then, his mother’s present age (in years) = x + 26.
3 years from now, Rohan’s age (in years) will be x + 3 and his mother’s age (in years) will be x + 29.
The product of their ages (in years) 3 years from now is given to be 360.
Hence, (x + 3)(x + 29) = 360
∴ x² + 32x + 87 – 360 = 0
∴ x² + 32x – 273 = 0 is the required quadratic equation to find the present ages (in years) of Rohan (x) and his mother (x + 26).

4. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
Let the usual uniform speed of the train be x km/h.
Now, Time = \(\frac{\text { distance }}{\text { speed }}\)
∴ Time required to cover 480 km distance at usual speed = t₁ = \(\frac{480}{x}\) hours
If the speed is 8 km/hour less, the new speed would be (x – 8) km/hour.
∴ Time required to cover 480 km distance at new speed = t₂ = \(\frac{480}{x-8}\) hours
Now, the time required at new speed is 3 hours more than the usual time.
∴ t₂ = t₁ + 3
∴ \(\frac{480}{x-8}=\frac{480}{x}+3\)
∴ 480x = 480(x – 8) + 3x(x – 8)
(Multiplying by x(x – 8))
∴ 480x = 480x – 3840 + 3x² – 24x
∴ 0 = 3x² – 24x – 3840
∴ x² – 8x – 1280 = 0 is the required quadratic equation to find the usual speed (x km/h) of the train.

Jharkhand Board JAC Class 10 Sanskrit Solutions Shemushi Chapter 12 अन्योक्तयः Textbook Exercise Questions and Answers.

JAC Board Class 10th Sanskrit Solutions Shemushi Chapter 12 अन्योक्तयः

JAC Class 10th Sanskrit अन्योक्तयः Textbook Questions and Answers

प्रश्न 1.
एकपदेन उत्तरं लिखत (एक शब्द में उत्तर लिखिए)
(क) कस्य शोभा एकेन राजहंसेन भवति?
(किसकी शोभा एक राजहंस से होती है?)
उत्तरम् :
सरसः (सरोवर की)।

(ख) सरसः तीरे के वसन्ति?
(तालाब के किनारे कौन रहते हैं?)
उत्तरम् :
वकसहस्रम् (हजारों बगुले)।

(ग) कः पिपासः प्रियते?
(कौन प्यासा मर जाता है?)
उत्तरम् :
चातकः (पपीहा)।

(घ) के रसाल मुकुलानि समाश्रयन्ते ?
(आम की मंजरियों का आश्रय कौन लेते हैं ?)
उत्तरम् :
भृङ्गाः (भौरे)।

(ङ) अम्भोदाः कुत्र सन्ति? (बादल कहाँ हैं?)
उत्तरम् :
गगने (आकाश में)।

प्रश्न 2.
अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत
(नीचे लिखे प्रश्नों के उत्तर संस्कृत भाषा में लिखिए)
(क) सरसः शोभा केन भवति ?
(तालाब की शोभा किससे होती है ?)
उत्तरम् :
सरसः शोभा एकेन एव राजहंसेन भवति।
(तालाब की शोभा एक ही राजहंस से होती है।)

(ख) चातकः किमर्थं मानी कथ्यते ?
(चातक किस अर्थ में स्वाभिमानी है?)
उत्तरम् :
चातकः तृषितः मरणम् आप्नोति परञ्च सर्वान् वारिदान् न याचते, केवलं पुरन्दरं याचते।
(चातक प्यासा मर जाता है, लेकिन अन्य बादलों से नहीं माँगता, केवल इन्द्र से माँगता है।)

(ग) मीनः कदा दीनां गतिं प्राप्नोति ?
(मछली कब दीन गति को प्राप्त होती है ?)
उत्तरम् :
यदा सरः सङ्कोचमञ्चति तदा मीन: दीनां गतिं प्राप्नोति।
(जब तालाब सूख जाता है तब मछली दीन दशा को प्राप्त होती है।)

(घ) कानि पूरयित्वा जलदः रिक्तः भवति ?
(किन्हें भरकर बादल खाली हो जाता है ?)
उत्तरम् :
नानानदीनदशतानि पूरयित्वा जलदः रिक्तः भवति।
(अनेक नदी और सैकड़ों नदों को भरकर बादल खाली हो जाता है।)

(ङ) वृष्टिभिः वसुधां के आर्द्रयन्ति ?
(वर्षा से धरती को कौन गीला कर देते हैं ?)
उत्तरम् :
अम्भोदाः वृष्टिभिः वसुधां आर्द्रयन्ति।
(बादल वर्षा से धरती को गीला कर देते हैं।)

प्रश्न 3.
अधोलिखितवाक्येष रेखाकितपदानि आधुत्य प्रश्ननिर्माणं करुत –
(निम्नलिखित वाक्यों में रेखांकित पदों के आधार पर प्रश्न निर्माण कीजिए)
(क) मालाकारः तोयैः तरोः पुष्टिं करोति।
(माली जल से वृक्ष का पोषण करता है।)
उत्तरम् :
मालाकार: कैः तरोः पुष्टिं करोति ?
(माली किनसे वृक्ष का पोषण करता है ?)
(ख) भृङ्गाः रसालमुकुलानि समाश्रयन्ते।
(भौरे आम के बौर का आश्रय लेते हैं।)
उत्तरम् :
भृङ्गाः कानि समाश्रयन्ते ?
(भौरे किनका आश्रय लेते हैं ?)
(ग) पतङ्गाः अम्बरपथम् आपेदिरे।
(पक्षियों ने आकाश मार्ग पाया।)
उत्तरम् :
के अम्बरपथं आपेदिरे ?
(किन्होंने आकाश मार्ग पाया ?)
(घ) जलदः नानानदीनदशतानि पूरयित्वा रिक्तोऽस्ति।
(बादल अनेक नदी और सैकड़ों नदों को भरकर खाली है।)
उत्तरम् :
कः नानानदीनदशतानि पूरयित्वा रिक्तोऽस्ति ?
(कौन अनेक नदी और सैकड़ों नदों को भरकर खाली है ?)

(ङ) चातकः वने वसति। (पपीहा वन में रहता है।)
उत्तरम् :
चातकः कुत्र वसति ? (पपीहा कहाँ रहता है ?)

प्रश्न 4.
अधोलिखितयो श्लोकयोः भावार्थ संस्कृतभाषया लिखत –
(निम्नलिखित श्लोकों का भावार्थ संस्कृत भाषा में लिखिये)
(अ) तोयैरल्पेरपि ………… वारिदेन।।
उत्तरम् :
भावार्थ – सुयोग्यः समर्थः एव व्यक्ति कार्य सम्यग् रूपेण सम्पादयितुं शक्नोति। यथा हि कवि पण्डित राज कथयति-‘रे उद्यानपाल! यं वृक्षं त्वं सकरुणं प्रचण्ड ग्रीष्म? अल्पैः जलै, अर्थात् आवश्यकतानुसारमसिञ्चः, एवम् असौ वृक्षः शनैः पालितः, किं तत् कार्यम् समर्थोऽसन् वृष्टिः धारा प्रवाहेण प्रभूतेन जलेन कर्तुं शक्नोति।

यथावश्यकमेव पोषणम् अपेक्ष्यते।’ (सुयोग्य और समर्थ व्यक्ति ही कार्य सुचारु रूप से सम्पन्न कर सकता है। जैसा कि कवि पण्डितराज कहते हैं- ‘हे माली! जिस वृक्ष को तुमने करुणा के साथ प्रचण्ड ग्रीष्म ऋतु में आवश्यकतानुसार थोड़ा-थोड़ा पानी देकर पाला था, क्या इस कार्य में समर्थ होते हुये भी वर्षा (अनावश्यक) बहुत से धारा प्रवाह जल से कर सकता है। सभी यथावश्यक पोषण की अपेक्षा करते हैं।)

(आ) रे रे चातक ………………….. दीनं वचः।

भावार्थ: – श्लोकेऽस्मिन् कवि चातकस्य माध्यमेन कथयति यत् सर्वे धनिकाः समृद्धाः वा जनाः दातारः न भवन्ति अतः यं कञ्चित् मानवं मा याचनां कुरु यथा हि कविः भर्तहरि कथयति – ‘रे रे चातक! सावहितः सन् शृणु मित्र। आकाशे अनेके मेघाः आयान्ति यान्ति च परञ्च तेषु केचनैव वर्षन्ति। एवं एव संसारे अनेके धनिका समृद्धाः च मानवा आयान्ति यान्ति च। तेषु केचन यच्छन्ति अन्ये तु व्यर्थमेव आत्मानं प्रदर्शयन्ति। अतोऽहं ब्रवीमि यत् त्वम यं कञ्चनं पश्यति तस्य सम्मुखं याचनाय मा गच्छ। सु दानभावोपेतं जनं ज्ञात्वा एवं याचनाय हस्तौ प्रसारय।

यथा सर्वे वारिदाः चातकाय स्वाति बिन्दुं न दातुं शक्नोति तथैव सर्वे जनाः याचकाय दान न दातुं शक्नुवन्ति। (इस श्लोक में कवि चातक के माध्यम से कहता है कि सभी धनवान या समद्ध लोग दानदाता नहीं होते। अतः जिस किसी मानव से मत याचना करो। जैसा कि कवि भर्तृहरि कहता है-‘ओ चातक! सावधान होकर सुन। मित्र! आकाश में अनेक मेघ आते हैं और चले जाते हैं। उनमें से कुछ ही देते हैं अन्य तो व्यर्थ ही अपना दिखावा करते हैं। अतः मैं चाहता हूँ कि तुम जिस किसी को देखो उसी के सामने याचना के लिये मत जाओ। (अच्छे दान भाव से युक्त व्यक्ति को जानकर याचना के लिये हाथ फैलाओ। जैसे सभी बादल चातक को स्वाति की बूंद नहीं दे सकते हैं वैसे ही सभी लोग याचक को दान नहीं दे सकते हैं।)

प्रश्न 5.
अधोलिखित श्लोकयोः अन्वयं लिखत। (निम्नलिखित श्लोकों का अन्वय लिखो)
(अ) आपेदिरे …………. कतमां गतिमभ्युपैति।
अन्वयः – पतङ्गाः परितः अम्बरपथम् आपेदिरे, भृङ्गाः रसाल मुकलानि समाश्रयन्ते। सरः त्वयि सङ्कोचम् अञ्चति, हन्त दीन-दीनः मीनः नु कतमां गतिम् अभ्युपैतु।

(आ) आश्वास्य ………….. सैवतवोत्तमा श्रीः।।
अन्वयः – तपनोष्णतप्तम् पर्वतकुलम् आश्वास्य उदाम दावविधुराणि काननानि च आश्वास्य नाना नदी गतानि पूरयित्वा च हे जलद। यत् रिक्तः असि तव सा एव उत्तमा श्रीः।

प्रश्न 6.
उदाहरणमनुसृत्य सन्धिं/सन्धिविच्छेदं वा कुरुत –
(उदाहरण के अनुसार सन्धि/सन्धि-विच्छेद कीजिए)
(i) यथा – अन्य + उक्तयः = अन्योक्तयः।
(क) …………. + ………….. = निपीतान्यम्बूनि।
(ख) …………. + उपकारः = कृतोपकारः।
(ग) तपन + …………… = तपनोष्णतप्तम्।
(घ) तव + उत्तमा = …………………।
(ङ) न + एतादृशाः = …………………।
उत्तरम् :
(क) निपीतानि + अम्बूनि = निपीतान्यम्बूनि।
(ख) कृत + उपकारः = कृतोपकारः।
(ग) तपन + उष्णतप्तम् = तपनोष्णतप्तम्।
(घ) तव + उत्तमा = तवोत्तमा।
(ङ) न + एतादृशाः = नैतादृशाः।

(ii) यथा – पिपासितः + अपि = पिपासितोऽपि।
(क) ……………. + ……………… = कोऽपि।
(ख) ……………. + ………………. = रिक्तोऽसि।
(ग) मीनः + अयम् = ………………।
(घ) सर्वे + आदि = …………
उत्तरम् :
(क) कः + अपि = कोऽपि।
(ख) रिक्तः + असि = रिक्तोऽसि।
(ग) मीनः + अयम् = मीनोऽयम्।
(घ) सर्वे + ऽपि = ………………..

(iii) यथा – सरस: + भवेत् = सरसोभवेत्।
(क) खगः + मानी – ……………।
(ख) …………. + नु = मीनो नु।
(ग) पिपासितः + वा = ………..।
(घ) …………. + ………. = पुरुतोमा।
उत्तरम् :
(क) खगः + मानी = खगो मानी।
(ख) मीनः + नु = मीनो नु।
(ग) पिपासितः + वा = पिपासितोवा।
(घ) पुरतः + मा = पुरुतोमा।

(iv) यथा – मुनिः + अपि = मुनिरपि।
(क) तोयैः + अल्पैः = …………..।
(ख) ………… + अपि = अल्पैरपि।
(ग) तरोः + अपि = ………………।
(घ) ………. + आर्द्रचन्ति = वृष्टिमिरार्द्रयन्ति।
उत्तरम् :
(क) तोयैः + अल्पैः = तोयैरल्पैः
(ख) अल्पैः + अपि = अल्पैरपि
(ग) तरोः + अपि = तरोरपि।
(घ) वृष्टिमिः + आर्द्रयन्ति = वृष्टिमिरार्द्रयन्ति।

प्रश्न 7.
उदाहरणमनुसृत्य अधोलिखितैः विग्रहपदैः समस्तपदानि रचयत –
(उदाहरण के अनुसार निम्नलिखित विग्रह पदों से समस्त पद बनाइए)

JAC Class 10th Sanskrit अन्योक्तयः Important Questions and Answers

शब्दार्थ चयनम् –

अधोलिखित वाक्येषु रेखांकित पदानां प्रसङ्गानुकूलम् उचितार्थ चित्वा लिखत –

प्रश्न 1.
एकेन राजहंसेन या शोभा सरसो भवेत्।
(अ) तडागस्य
(ब) कटु
(स) सरसः
(द) जलदः
उत्तरम् :
(अ) तडागस्य

प्रश्न 2.
न सा. बकसहस्रेण परितस्तीरवासिना।
(अ) भवेत्
(ब) अभितः
(स) शृगालः
(द) भवति
उत्तरम् :
(ब) अभितः

प्रश्न 3.
भुक्ता मृणालपटली भवता निपीता।
(अ) भवता
(ब) युक्ता
(स) खादिता
(द) अम्बूनि
उत्तरम् :
(स) खादिता

प्रश्न 4.
न्यम्बूनि यत्र नलिनानि निषेवितानि।
(अ) राजहंस
(ब) सरोवरस्य
(स) भविता
(द) सेवितानि
उत्तरम् :
(द) सेवितानि

प्रश्न 5.
तोवैरल्यैरपि करुणया भीमभानौ निदाघे।
(अ) ग्रीष्मकाले
(ब) भानवेः
(स) निर्दय
(घ) भवता
उत्तरम् :
(अ) ग्रीष्मकाले

प्रश्न 6.
व्यरचि भवता या तरोरस्य पुष्टिः।
(अ) विश्वतः
(ब) कृता
(स) जनयितुम्
(द) शक्या
उत्तरम् :
(ब) कृता

प्रश्न 7.
भृङ्गा रसालमुकुलानि समाश्रयन्ते।
(अ) आपेदिरे
(ब) समाश्रयन्ते
(स) द्विरेफा
(द) दीनदीनः
उत्तरम् :
(स) द्विरेफा

प्रश्न 8.
एक एव खगो मानी वने वसति चातकः।
(अ) पिपासितः
(ब) पुरन्दरम्
(स) याचते
(द) स्तोककः
उत्तरम् :
(द) स्तोककः

प्रश्न 9.
यं यं पश्यसि तस्य तस्य पुरतो मा ब्रूहि दीनं वचः।
(अ) वद
(ब) बहवः
(स) आर्द्रयन्ति
(द) पश्यसि
उत्तरम् :
(अ) वद

प्रश्न 10.
नानानदीनदशतानि च पूरयित्वा।
(अ) पूर्णिमाया
(ब) पूरणी:
(स) आश्वास्य
(द) पूर्णं कृत्वा
उत्तरम् :
(स) आश्वास्य

संस्कृतमाध्यमेन प्रश्नोत्तराणि –

एकपदेन उत्तरत (एक शब्द में उत्तर दीजिए)

प्रश्न 1.
राजहंसोऽत्र कस्य प्रतीकः ?
(राजहंस यहाँ किसका प्रतीक है ?)
उत्तरम् :
सज्जनस्य (सज्जन का)।

प्रश्न 2.
कविः ‘बकसहस्रम्’ कं कथयति ?
(कवि ‘बकसहस्रम्’ किसे कहता है ?)
उत्तरम् :
मूर्खान् (मूरों को)।

प्रश्न 3.
मालाकारेण तरोः पुष्टिः कैः कृता ?
(माली ने वृक्ष का पोषण किनसे किया ?)
उत्तरम् :
तोयैः (जल से)।

प्रश्न 4.
मालाकारः कथं तोयैः वृक्षस्य पुष्टिः व्यरचि ?
(माली ने किस प्रकार पानी से वृक्ष की पुष्टि की ?)
उत्तरम् :
करुणया (करुणा से)।

प्रश्न 5.
सङ्कचिते सरोवरे पतङ्गाः कुत्र आपेदिरे ?
(सरोवर के सूख जाने पर पक्षी कहाँ गये ?)
उत्तरम् :
अम्बरपथम् (आकाश मार्ग में)।

प्रश्न 6.
कीदृशः चातकः पिपासितो म्रियते ?
(कैसा चातक प्यासा मरता है ?)
उत्तरम् :
मानी (स्वाभिमानी)।

प्रश्न 7.
कविः कं सम्बोध्य श्लोकं अरचयत् ?
(कवि किसको संबोधित करके श्लोक की रचना करता है?)
उत्तरम् :
मेघः (बादल)।

प्रश्न 8.
काननानि कः आश्वास्यति ?
(जंगलों को कौन आश्वस्त करता है ?)
उत्तरम् :
जलदः (बादल)।

प्रश्न 9.
कविः मित्रः इति शब्देन के सम्बोधयति ?
(कवि ‘मित्र’ इस शब्द से किसको संबोधित करता है?)
उत्तरम् :
चातकः (पपीहा को)।

प्रश्न 10.
केचिद् अम्भोदाः कथं गर्जन्ति ?
(कुछ बादल कैसे गर्जते हैं ?)
उत्तरम् :
वृथा (व्यर्थ ही)।

प्रश्न 11.
एकेन राजहंसेन कस्य शोभा भवेत् ?
(एक हंस से किसकी शोभा होती है?)
उत्तरम् :
सरसः (तालाब की)।

प्रश्न 12.
सरसः तीरवासिनः के सन्ति ?
(तालाब के किनारे वास करने वाले कौन हैं ?)
उत्तरम् :
बकसहस्रम् (हजारों बगुले)।

प्रश्न 13.
मृणालपटली केन भुक्ता ?
(कमलनालों का समूह किसके द्वारा खाया गया है ?)
उत्तरम् :
राजहंसेन (राजहंस के द्वारा)।

प्रश्न 14.
राजहंसेन कानि निपीतानि ?
(राजहंस ने किन्हें पीया ?)
उत्तरम् :
अम्बूनि (जल)।

प्रश्न 15.
तरोः पुष्टिः केन कृतः ?
(वृक्ष का पोषण किसके द्वारा किया गया है ?)
उत्तरम् :
मालाकारेण (माली द्वारा)।

प्रश्न 16.
मालाकारेण तरोः कदा पुष्टिः कृता ?
(माली ने वृक्ष का पोषण कब किया ?)
उत्तरंम् :
निदाघे (गर्मी में)।

प्रश्न 17.
सङ्कुचिते सरोवरे परितः अम्बरपथं कः आपेदिरे ?
(सरोवर के सूख जाने पर चारों ओर आकाशमार्ग को कौन प्राप्त कर लेते हैं ?)
उत्तरम् :
पतङ्गाः (पक्षी)।

प्रश्न 18.
रसालमुकुलानि के समाश्रयन्ते ?
(आम की मजरी को आश्रय कौन लेते हैं ?)
उत्तरम् :
भृङ्गाः (भौरे)।

प्रश्न 19.
एकः एव मानी वने कः वसति ?
(एक ही स्वाभिमानी वन में कौन निवास करता है ?)
उत्तरम् :
खगः (पक्षी)।

प्रश्न 20.
पुरन्दरात् कः याचते ?
(इन्द्र से कौन माँगता है ?)
उत्तरम् :
चातकः (पपीहा)।

प्रश्न 21.
कः नदी नद शतानि पूरयति ?
(कौन सैकड़ों नद-नदियों को भर देता है ?)
उत्तरम् :
जलदः (बादल)।

प्रश्न 22.
तपनोष्णतप्तम् किम् ?
(सूर्य की गर्मी से कौन तपी है ?)
उत्तरम् :
पर्वतकुलम् (पर्वतों के समूह)।

प्रश्न 23.
के वृष्टिभः वसुधां आर्द्रयन्ति ?
(कौन बरसकर धरती को गीला कर देते हैं ?)
उत्तरम् :
अम्भोदाः (बादल)।

प्रश्न 24.
कीदृशं वचः मा ब्रूहिः ?
(कैसे वचन मत बोलो?)
उत्तरम् :
दीनंवचः (दीन वचन)।

पूर्णवाक्येन उत्तरत (पूरे वाक्य में उत्तर दीजिए)

प्रश्न 25.
कृतोपकारः कः भविष्यति ?
(उपकार करने वाला कौन होगा ?)
उत्तरम् :
कृतोपकारः राजहंसः भविष्यति।
(उपकार करने वाला राजहंस होगा।)

प्रश्न 26.
धारासारान् कः विकिरति।
(जलधाराओं को कौन बरसाता है ?)
उत्तरम् :
धारासारान् वारिदः विकिरति।
(जलधाराओं को बादल बरसाता है।)

प्रश्न 27.
चातकः मानी कथम् ?
(चातक मानी कैसे है ?)
उत्तरम् :
चातकः पिपासितो म्रियते पुरन्दरम् वा याचते।
(चातक या तो प्यासा मरता है या इन्द्र से याचना करता है।)

प्रश्न 28.
जलदः कान् पूरयति ? (बादल किनको भरता है?)
उत्तरम :
जलदः नानानदीनदशतानि पूरयति।
(बादल अनेक नदी और सैकड़ों नदों को भरता है।)

प्रश्न 29.
कविः श्रोतारं किं कारणाद् वर्जयति?
(कवि श्रोताओं को किस कारण रोकता है?)
उत्तरम् :
कवि श्रोतारं वर्जयति-मा ब्रूहि दीनं वचः
(कवि सुनने वालों को रोकता है–किसी के सामने दीन वचन मत बोलो)।

प्रश्न 30.
सरसः शोभा केन न भवति ?
(तालाब की शोभा किससे नहीं होती ?)
उत्तरम् :
सरसः शोभा परितः तीरवासिना बकसहस्रेण न भवति।
(तालाब की शोभा चारों ओर किनारे पर बैठे हजारों बगलों से नहीं होती।)

प्रश्न 31.
राजहंसेन कानि निषेवितानि ?
(राजहंस द्वारा किनका सेवन किया गया ?)
उत्तरम् :
राजहंसेन नलिनानि निषेवितानि।
(राजहंस ने कमलनालों का सेवन किया।)

प्रश्न 32.
मालाकार: केन प्रकारेण तरोः पुष्टिं करोति ?
(माली किस प्रकार से वृक्ष की पुष्टि करता है ?)
उत्तरम् :
मालाकार: निदाघे अल्पैः तोयैः अपि करुणया तरोः पुष्टिं करोति।
(माली गर्मी में थोड़े जल से भी करुणा के साथ वृक्ष का पोषण करता है।)

प्रश्न 33.
सरोवरे सङ्कुचिते भृङ्गाः कानि समाश्रयन्ते ?
(सरोवर के सूख जाने पर भौरे किनका आश्रय ले लेते हैं?)
उत्तरम् :
सरोवरे सङ्कुचिते भृङ्गाः रसालमुकुलानि समाश्रयन्ते।
(सरोवर के सूख जाने पर भौरे आम की मंजरी का आश्रय ले लेते हैं।)

प्रश्न 34.
वने कः वसति ?
(वन में कौन रहता है ?)
उत्तरम् :
वने एकः मानी खगः चातकः वसति।
(वन में एक स्वाभिमानी पक्षी चातक रहता है।)

प्रश्न 35.
जलदः कीदृशानि काननानि आवश्स्य रिक्तो भवति ?
(बादल वनों को किस प्रकार आश्वस्त करके जलहीन (खाली) हो जाता है?)
उत्तरम् :
जलदः रिक्तोभवति उद्दामदावविधुराणि काननानि आश्वस्तं करोति।
(बादल खाली होकर दावाग्नि से नष्ट हुए वनों को आश्वस्त करता है।)

प्रश्न 36.
कवि के सम्बोध्य श्लोकं कथयति ?
(कवि श्लोक किसको सम्बोधित करके कहता है?)
उत्तरम् :
कवि चातकः सम्बोध्य श्लोक कथयति।
(कवि श्लोक चातक को सम्बोधित करके कहता है।

अन्वय-लेखनम् –

अधोलिखितश्लोकस्यान्वयमाश्रित्य रिक्तस्थानानि मञ्जूषातः समुचितपदानि चित्वा पूरयत।
(नीचे लिखे श्लोक के अन्वय के आधार पर रिक्तस्थानों की पूर्ति मंजूषा से उचित पद चुनकर कीजिए।)

1. एकेन ……….. ………… तीरवासिना।। .
मञ्जूषा – तीरवासिना, राजहंसेन, सा, शोभा।
एकेन (i) …………. सरस: या (ii) …………. भवेत् परितः (iii) …………. बकसहस्रेण (iv) …………. (शोभा) न (भवति)।
उत्तरम् :
(i) राजहंसेन (ii) शोभा (iii) तीरवासिना (iv) सा।

2. भुक्ता …………………. कृतोपकारः।।
मञ्जूषा – निषेवितानि, अम्बूनि, मृणालपटली, कृतोपकारः।
यत्र भवता (i) ………… भुक्ता, (ii) ………… निपीतानि नलिनानि (iii) ………… रे राजहंस ! तस्य सरोवरस्य केन कृत्येन (iv) …………. भविता असि, वद।
उत्तरम् :
(i) मृणालपटली (ii) अम्बूनि (iii) निषेवितानि (iv) कृतोपकारः।

3. तोयैरल्पैरपि………………………. वारिदेन।।
मञ्जूषा – प्रावृषेण्येन, भीमभानौ, विकिरता, करुणया।
हे मालाकार ! (i) …………. निदाघे अल्पैः तोयैः अपि भवता (ii) …………. अस्य तरोः या पुष्टिः व्यरचि। वारां (iii) …………. विश्वतः धारासारान् अपि (iv) …………. वारिदेन इह जनयितुं सा किं शक्या।
उत्तरम् :
(i) भीमभानौ (ii) करुणया (iii) प्रावृषेण्येन (iv) विकिरता।

4. आपेदिरेऽम्बरपथं ………………… गतिमभ्युपैतु।।
मञ्जूषा – कतमां, भृङ्गाः, परितः, अञ्चति। (सङ्कुचिते सरोवरे)
पतङ्गाः (i) …………. अम्बरपथम् आपेदिरे, (ii) …………. रसालमुकुलानि समाश्रयन्ते। सरः त्वयि सङ्कोचम् (iii) …………., हन्त दानदान: मीनः नु (iv) ………….गतिम् अभ्युपतु।।
उत्तरम् :
(i) परितः (ii) भृङ्गाः (iii) अञ्चति (iv) कतमां।

5. एक एव खगो ……………………….. पुरन्दरम्।।
मञ्जूषा – पुरन्दरं, चातकः, एव, पिपासितः।
एक: (i) ……….. मानी खगः (ii) ……….. वने वसति वा (iii) ……….. म्रियते (ii) ………… याचते।।
उत्तरम् :
(i) एव (ii) चातकः (iii) पिपासितः (iv) पुरन्दरं।।

6. आश्वास्य पर्वतकुलं ……….. …… तवोत्तमा श्रीः।।
मञ्जूषा – काननानि, पर्वतकुलम्, उत्तमा, जलद।
तपनोष्णतप्तम् (i) ……….. आश्वास्य उद्दामदावविधुराणि (ii) ……….. च (आश्वास्य) नानानदीनदशतानि पूरयित्वा च हे (ii) ……….. ! यत् रिक्तः असि तव सा एव (ii) ……….. श्रीः।
उत्तरम् :
(i) पर्वतकुलम् (ii) काननानि (iii) जलद (iv) उत्तमा।

7. रे रे चातक ! ……………………….. दान …….. दीनं वचः।।
मञ्जूषा – एतादृशाः, अम्भोदाः, श्रूयताम, आर्द्रयन्ति।
रे रे मित्र चातक ! सावधानमनसा क्षणं (i) …………. गगने हि बहवः (ii) …………. सन्ति, सर्वेऽपि (iii) … ” ………. न (सन्ति) केचित् वसुधां वृष्टिभिः (iv) …………. केचित् वृथा गर्जन्ति, यं यं पश्यसि तस्य तस्य पुरतः दीनं वचः मा ब्रूहि।
उत्तरम् :
(i) श्रूयताम् (ii) अम्भोदाः (iii) एतादृशाः (iv) आर्द्रयन्ति।

प्रश्ननिर्माणम् –

अधोलिखित वाक्येषु स्थूलपदमाधृत्य प्रश्ननिर्माणं कुरुत –

1. सरसः शोभा राजहंसेन भवेत्.। (तालाब की शोभा राजहंस से होनी चाहिए।)
2. राजहंसेन सरोवरस्य नलिनानि निषेवितानि। (राजहंस द्वारा सरोवर के कमलनालों का सेवन किया गया।)
3. भृङ्गाः रसालमुकुलानि समाश्रयन्ते। (भौरे आम के बौर का आश्रय लेते हैं।)
4. पतङ्गाः परितः अम्बरपथं आपेदिरे। (पक्षी चारों ओर आकाश में पहुँच गए।)
5. सरसि संकोचम् अञ्चति मीन: दु:खीभवति। (तालाब के सूख जाने पर मछली दःखी होती है।)
6. मानी चातकः वने वसति। (स्वाभिमानी चातक वन में रहता है।)
7. चातक: पुरन्दरं याचते। (पपीहा इन्द्र से याचना करता है।)।
8. जलदः तपनोष्णतप्तं पर्वतकुलम् आश्वासयति। (बादल सूर्य से सन्तप्त पर्वत समूहों को आश्वस्त करता है।) .
9. जलदः नानानदीनदशतानि पूरयित्वा रिक्तो भवति। (बादल सैकड़ों नदी-नदों को भरकर रिक्त होता है।)
10. वारिदाः वृष्टिभिः वसुधाम् आर्द्रयन्ति। (बादल वर्षा से धरती को गीला करते हैं।)
उत्तराणि :
1. सरसः शोभा केन भवेत् ?
2. राजहंसेन सरोवरस्य कानि निषेवितानि ?
3. भृङ्गाः कानि समाश्रयन्ते ?
4. पतङ्गाः परितः कम् आपेदिरे ?
5. सरसि सङ्कोचमञ्चति कः दुःखी भवति ?
6. मानी चातकः कुत्र वसति ?
7. चातकः कम् याचते ?
8. जलदः के आश्वासयति ?
9. जलदः कानि पूरयित्वा रिक्तो भवति ?
10. वारिदाः काभिः वसुधाम् आर्द्रयन्ति ?

भावार्थ-लेखनम् –

अधोलिखित पद्यांश संस्कृते भावार्थ लिखत –

1. एकेन राजहंसेन या ………………………. परितस्तीरवासिना।।
भावार्थ – एकेन मरालेन सरोवरस्य सौन्दर्य स्मात् परञ्च अभितः तट उषितैः सहस्रवकैः सा शोभा न सम्भवति।

2. भुक्ता मृणालपटली भवता …………………………… भवितासि कृतोपकारः।।
भावार्थ – यस्मिन् स्थाने त्वया श्रीमता कमलनाल समूह खादितः नि:शेषाणि जलानि पीतानि कमलानां सेवनं कृतम् रे मराल! अमुष्य तडागस्य केन कार्येण तस्य हितकारकः भविष्यसि (कथं तस्योपकारं करिष्यसि) कथय कथं भविष्यति।

3. तोयैरल्पैरपि करुणया ……………………….. विकिरता विश्वतो वारिदेन।।
भावार्थ – हे उद्यान पाल! भयंकर प्रचण्ड सूर्ये प्रतप्ते ग्रीष्मौ किञ्चित जलेन अपि त्वया अनुकम्पया त्वया एतस्य वृक्षस्य यत्पोषणं त्वया कृतम् किं वर्षा कालिकेन सर्वतः जलधारा वर्षयता जलदेन कर्तुं शक्यते।

4. आपेदिरेऽम्बरपथं परितः …………………………………. कतमां गतिमभ्युपैतु।।
भावार्थ – शुष्के सरसि खगाः सर्वतः आकाशमार्गम् प्राप्नुवन्ति, भ्रमराः रसालानां मञ्जरीणां शरणं गृह्णन्ति। हे सरोवर! त्वयि जलाभावेसति हा अतिदीनाः मत्स्याः कां स्थिति प्राप्स्यन्ति अर्थात् तेषां न कोऽपि आश्रयः।

5. एक एव खगो मानी ……………………….. याचते वा पुरन्दरम्।।
भावार्थ – एकः एव स्वाभिमानी पक्षी सारङ्ग स्तोकको वा कानने निवसति, सः तृषितः मरणम् आप्नोति अन्यथा इन्द्रं देवराजमेव याचनां करोति। (सर्वान् अन्यान् वारिदान् न याचते)।

6. आश्वास्य पर्वतकुलं ………………. यज्जलद ! सैव तवोत्तमा श्रीः।।
भावार्थ – सूर्यस्य उष्णतया प्रतप्तम् गिरिणाम् समूहे विश्वासम् उत्पाद्य विश्वस्तं वा विधायः, उन्नत काष्ठ रहितानि वनानि च समाश्वास्य, अनेकाषां सरितां महानदानां शतानां च कलेवरान् पूर्ण विधाय हे वारिद् ! यत्त्वम् स्वकीयं सर्वं त्यक्त्वा जलहीनो जातोऽसि असावेव ते श्रेष्ठतमा शोभा तदैव ते महनीयता।

7. रे रे चातक ! सावधानमनसा ………………………………. मा ब्रूहि दीनं वचः।।
भावार्थ – हे सुहृद सारङ्ग! सावहितः सन् किञ्चित् कालम् आकर्णताम् यतः आकाशे अनेके वारिदाः सन्ति (विद्यन्ते) परञ्च निः शेषाः एव न वर्तन्ते एवं विधा। केचिद् तु धराम् वर्षायाः जलेन सिञ्चन्ति यावत् केचन तु व्यर्थमेव गर्जनां कुर्वन्ति अतः त्वं यं यं वारिदं ईक्षसे अमुष्य अमुष्य सम्मुखे याचनायाः करुणं वचनं न वद अर्थात् तं न याचस्व।

अधोलिखितानां सूक्तीनां भावबोधनं सरलसंस्कृतभाषया लिखत –
(निम्नलिखित सूक्तियों का भावबोधन सरल संस्कृत भाषा में लिखिए-)

(i) एकेन राजहंसेन या शोभा सरसो भवेत् न सा बकसहस्रेण।
भावार्थः – यथा एकेन एव हंसेन सरोवरस्य सौन्दर्यं वर्धते न तु सहस्रः बकैः तथैव एकेन एव सज्जनेन पण्डितेन सभायाः शोभा वर्धते न च सहस्रः मूर्खः। (जैसे एक ही हंस से सरोवर का सौन्दर्य बढ़ जाता है न कि हजारों बगुलों से, इसी प्रकार एक ही सज्जन या विद्वान् से सभा की शोभा बढ़ जाती है न कि हजारों मूों से।)

(ii) एक एव खगो मानी वने वसति चातकः।
पिपासितो वा म्रियते याचते वा पुरन्दरम्।।
भावार्थ: – यथा स्वाभिमानी चातकः वने निवसन्नपि तृषया मृत्यु प्राप्नोति परञ्च स्वातिबिन्दुं विना अन्यं जलं न पिबति तथैव धीराः जनाः अपि मरणासन्ने अपि कस्माद् अपि न याचन्ते, ते परमात्मानम् एव याचन्ते। (जिस प्रकार स्वाभिमानी चातक वन में रहते हुए भी प्यास से मृत्यु को प्राप्त हो जाता है परन्तु स्वाति की बूँद के अतिरिक्त अन्य जल नहीं पीता, उसी प्रकार धीर पुरुष मरने की स्थिति में भी किसी से माँगते नहीं, वे परमात्मा से ही याचना करते हैं।)

(iii) यं यं पश्यसि तस्य तस्य पुरतो माहदानं वचः।
भावार्थ: – कविः चातकस्य व्याजेन मानवाय उपदिशति यत् संसार बहवः धनाढ्याः सन्ति परन्तु ते सर्वे उदारहदयं दातारः न भवन्ति.अत: यं यं पश्यसि त त मा याचे। (कवि बातक के बहाने मनुष्य को उपदेश देता है कि संसार में बहुत से धनवान् हैं परन्तु सभी उदार हृदय दाता नहीं हैं अतः जिस-जिस को देखो उसी-उसी से मत माँगो।)

अन्योक्तयः Summary and Translation in Hindi

पाठ-परिचय – अन्योक्ति का अर्थ है- अन्य (दूसरे) को लक्ष्य करके कही गई बात अर्थात् किसी की प्रशंसा अथवा निन्दा प्रत्यक्ष रूप से कहने की अपेक्षा किसी अन्य को माध्यम बनाकर कहना। कवियों की ऐसी अभिव्यक्ति को ही अन्योक्ति कहते हैं। ये उक्तियाँ (कथन) अत्यन्त मार्मिक होती हैं जो सीधे लक्ष्य का भेदन करती हैं। प्रस्तुत पाठ में ऐसी ही सात अन्योक्तियाँ विविध ग्रन्थों से संकलित की गई हैं जिनमें राजहंस, कोकिल, मेघ, मालाकार, सरोवर तथा चातकं के माध्यम से मानव को परोपकार आदि सवृत्तियों एवं सत्कर्मों में प्रवृत्त होने के लिए प्रेरित किया गया है। पाठ में संकलित अन्योक्तियों में छठी अन्योक्ति महाकवि माघ के ‘शिशुपालवध’ महाकाव्य से तथा सातवीं अन्योक्ति महाकवि भर्तहरि के ‘नीतिशतक’ से ली गई है तथा शेष पाँच पण्डितराज जगन्नाथ के ‘भामिनीविलास’ के अन्योक्ति भाग से संकलित हैं।

मूलपाठः,अन्वयः,शब्दार्थाः, सप्रसंग हिन्दी-अनुवादः।

1 एकेन राजहंसेन या शोभा सरसो भवेत्।
न सा बकसहस्रेण परितस्तीरवासिना।।1।।

अन्वयः – एकेन राजहंसेन सरसः या शोभा भवेत्। परितः तीरवासिना बकसहस्रेण सा (शोभा) न (भवति)।

शब्दार्थाः – एकेन राजहंसेन = एकेन मरालेन (एक हंस से), सरसः = तडागस्य, सरोवरस्य (तालाब की), या शोभा = या सौन्दर्यवृद्धिः, श्रीः (जो शोभा), भवेत् = स्यात् (होनी चाहिए), परितः = अभितः (चारों ओर), तीरवासिना = तटे स्थितैः उषितैः (किनारे पर वास करने वाले), बकसहस्त्रेण = सहस्त्रैः बकैः, बकानां सहस्रेण (हजारों बगुलों से), सा = असौ शोभा (वह शोभा) न भवति = (नहीं होती है)।।

सन्दर्भ-प्रसङ्गश्च – यह पद्य हमारी पाठ्यपुस्तक शेमुषी के ‘अन्योक्तयः’ पाठ से लिया गया है। मूलतः यह श्लोक पण्डितराज जगन्नाथ कृत ‘भामिनी विलास’ ग्रन्थ से संकलित है। इस श्लोक में कवि पण्डितराज जगन्नाथ सज्जन-दुर्जन का भेद वर्णन करते हैं।

हिन्दी-अनुवादः – एक (ही) हंस से तालाब की जो शोभा होनी चाहिए, चारों ओर किनारे पर रहने वाले हजारों बगुलों से वह शोभा नहीं होती।

2. भुक्ता मृणालपटली भवता निपीता न्यम्बूनि यत्र नलिनानि निषेवितानि।
रे राजहंस ! वद तस्य सरोवरस्य, कृत्येन केन भवितासि कृतोपकारः।।2।।

अन्वयः – यत्र भवता मृणालपटली भुक्ता, अम्बूनि निपीतानि नलिनानि निषेवितानि। रे राजहंस ! तस्य सरोवरस्य केन कृत्येन कृतोपकारः भविता असि, वद।।

शब्दार्थाः – यत्र = यस्मिन् स्थाने (जहाँ), भवता = त्वया श्रीमता (आपके द्वारा), मृणालपटली = कमलनाल समूहः (कमलनालों का समूह), भुक्ता = खादिता (खाया गया, भोगा गया), अम्बूनि = जलानि (जल), निपीतानि = निःशेषेण पीतानि, सम्पूर्णरूपेण पीतानि, सम्यक् पीतानि (भलीभाँति पीया गया), नलिनानि = कमलानि (कमलों को), निषेवितानि = सेवितानि (सेवन किए गए), रे राजहंस! = रे मराल ! (अरे राजहंस!), तस्य = अमुष्य (उस), सरोवरस्य = तडागस्य, सरसः (सरोवर का, तालाब का), केन कृत्येन – केन कार्येण (किस कार्य से), कृतोपकारः = कृतः उपकारः येन सः सम्पादितोपकारः, हितकारकः (उपकार किया हुआ, प्रत्युपकार करने वाला), भविता असि = भविष्यति (होगा), वद = ब्रूहि, कथय (बोलो)।

सन्दर्भ-प्रसङ्गश्च – यह पद्य हमारी पाठ्यपुस्तक शेमुषी के ‘अन्योक्तयः’ पाठ से लिया गया है। मूलतः यह श्लोक पण्डितराज जगन्नाथ विरचित ‘भामिनी विलास’ ग्रन्थ से संकलित है। इसमें कवि राजहंस के बहाने मनुष्य को प्रत्युपकार करने हेतु प्रेरित करता है।

हिन्दी-अनुवादः – जहाँ आपके द्वारा कमलनालों का समूह खाया गया, जल भलीभाँति पिया गया, कमलों का सेवन किया गया। हे राजहंस! उस तालाब का किस कार्य से प्रत्युपकार करने वाले होंगे अर्थात् उसका प्रत्युपकार किस कार्य से करेंगे।

3. तोयैरल्यैरपि करुणया भीमभानौ निदाघे, मालाकार ! व्यरचि भवता या तरोरस्य पुष्टिः।
सा किं शक्या जनयितुमिह प्रावृषेण्येन वारां, धारासारानपि विकिरता विश्वतो वारिदेन।।3।।

अन्वयः – हे मालाकार! भीमभानौ निदाघे अल्पैः तोयैः अपि भवता करुणया अस्य तरोः या पुष्टिः व्यरचि। वारां प्रावृषेण्येन विश्वतः धारासारान् अपि विकिरता वारिदेन इह जनयितुम् सा (पुष्टि:) किं शक्या।।

शब्दार्थाः – हे मालाकार ! = हे उद्यानपाल, हे सक्कार ! (अरे माली!), भीमभानौ = भीमः भानुः यस्मिन् सः भीमभानुः, तस्मिन् अति तीक्ष्णांशुमति तपति (ग्रीष्मकाल में सूर्य के अत्यधिक तपने पर), निदाघे = ग्रीष्मकाले (ग्रीष्म काल में), अल्पैः तोयैः अपि = किञ्चित् जलेन अपि (थोड़े पानी से भी), भवता = त्वया (आपके द्वारा), करुणया = अनुकम्पया, दयया (करुणा द्वारा, करुणा के साथ, दया से), अस्य तरोः = एतस्य वृक्षस्य (इस वृक्ष की), या पुष्टिः = या पुष्टता, या वृद्धिः, यत्पोषणम् (जो पोषण), व्यरचि = कृता, कृतम्, रचना क्रियते (की जाती है, की गई), वाराम् = जलानां (जलों के), प्रावृषेण्येन = वर्षाकालिकेन (वर्षाकालिक, वर्षाकाल के), विश्वतः = सर्वतः (सभी ओर), धारासारान् अपि = धाराणाम् आसारा अपि (जलधाराओं के प्रवाहों को भी), विकिरता = जलं वर्षयता (जल बरसाते हुए), वारिदेन = जलदेन (बादल द्वारा), इह = अस्मिन् संसारे (इस संसार में), जनयितुम् = उत्पादयितुम् (उत्पन्न/पैदा करने के लिए), सा पुष्टिः = तत्पोषणम् (वह पुष्टि), किम् शक्या = अपि सम्भवा, सम्भवति (क्या सम्भव है)।

सन्दर्भ-प्रसङ्गश्च – यह पद्य हमारी शेमुषी पाठ्यपुस्तक के ‘अन्योक्तयः’ पाठ से लिया गया है। मूलतः यह पद्य पण्डितराज जगन्नाथ रचित ‘भामिनी विलास’ काव्य से संकलित है। इस पद्य में कवि माली के बहाने से एक अच्छे पालनकर्ता के लक्षण बताता है।

हिन्दी-अनुवादः – अरे माली ! सूर्य के अत्यधिक तपने वाले ग्रीष्मकाल में थोड़े पानी से भी आपके द्वारा करुणा के साथ इस वृक्ष का जो पोषण किया गया है (किया जाता है) वर्षा-कालिक जल की सभी ओर से जलधाराओं के प्रवाह से भी जल बरसाते हुए बादल के द्वारा इस संसार में उस पोषण को पैदा करने में समर्थ है क्या ?

4. आपेदिरेऽम्बरपथं परितः पतङ्गाः, भृङ्गा रसालमुकुलानि समाश्रयन्ते।
सङ्कोचमञ्चति सरस्त्वयि दीनदीनो, मीनो नु हन्त कतमां गतिमभ्युपैतु।।4।।

अन्वयः – (सङ्कुचिते सरोवरे) पतङ्गाः परितः अम्बरपथम् आपेदिरे, भृङ्गाः रसालमुकुलानि समाश्रयन्ते। सरः त्वयि सङ्कोचम् अञ्चति, हन्त दीनदीन: मीनः नु कतमां गतिम् अभ्युपैतु।।

शब्दार्थाः – [सङ्कुचिते सरोवरे (सरोवर के सूख जाने पर)] पतङ्गाः = खगाः (पक्षी), परितः = सर्वतः, (सभी ओर), अम्बरपथम् = आकाशमार्गम् (आकाश मार्ग को), आपेदिरे = प्राप्तवन्तः, प्राप्नुवन्ति (प्राप्त कर लिए/लेते हैं), भृङ्गाः = भ्रमराः, द्विरेफाः (भौरे, भँवरे), रसालमुकुलानि = रसालानाम् आम्राणां मुकुलानि मञ्जरीम मञ्जाः मञ्जरीणाम् (आम की मञ्जरियों को/का), समाश्रयन्ते = शरणं प्राप्नुवन्ति (आश्रय लेते हैं), सरः = हे सरोवर ! हे तडाग ! (हे तालाब !), त्वयि = ते, तव भवति (तेरे), सङ्कोचम् अञ्चति = सङ्कचिते सति गच्छति (तुम्हारे संकुचित हो जाने पर, सूखकर जल कम हो जाने पर), हन्त = खेदः (खेद है), दीनदीनः = अतिदीनः (बेचारा), मीनः नुः = मत्स्यः , मत्स्यगणः (मछलियाँ), कतमां गतिम् = कां स्थितिं (किस गति को), अभ्युपैतु = प्राप्नोतु (प्राप्त करें)।

सन्दर्भ-प्रसङ्गश्च – यह पद्य हमारी शेमुषी पाठ्यपुस्तक के ‘अन्योक्तयः’ पाठ से लिया गया है। मूलतः यह पद्य पण्डितराज जगन्नाथ कृत् ‘भामिनी विलास’ काव्य से संकलित है। इस पद्य में कवि तालाब के माध्यम से मानव को उसकी स्थिति से अवगत कराते हुये कहता है। जैसे तालाब के सूख जाने पर सभी जीव उसे अकेला छोड़कर चले जाते हैं। वैसे मनुष्य को भी स्वार्थ पूरा होने पर छोड़ जाते हैं।

हिन्दी-अनुवादः – (सरोवर के सूख जाने पर) पक्षी चारों ओर आकाश मार्ग को प्राप्त कर लेते हैं अर्थात् आकाश में उड़ जाते हैं। भौरे आम की मञ्जरी का आश्रय ले लेते हैं (प्राप्त कर लेते हैं) हे तालाब ! तुम्हारे संकुचित हो जाने पर (सूख जाने पर) खेद है बेचारी मछलियाँ किस गति को प्राप्त करें (करेंगी)।

5. एक एव खगो मानी वने वसति चातकः।
पिपासितो वा म्रियते याचते वा पुरन्दरम्।।5।।

अन्वयः – एकः एव मानी खग: चातक: वने वसति वा पिपासितः म्रियते पुरन्दरं याचते।।

शब्दार्थाः — एक एव मानी = एक एव स्वाभिमानी (एक ही स्वाभिमानी), खगः = पक्षी, अण्डजः (पक्षी), चातकः = स्तोककः, सारङ्गः (पपीहा), वने = कानने (वन में), वसति = निवसति (रहता है), वा = अथवा (अथवा), पिपासितः = तृषितः (प्यासा), म्रियते = मरणं प्राप्नोति, मृत्युं लभते, मृत्युं वृणोति (मर जाता है), पुरन्दरम् = इन्द्रम् (इन्द्र से), याचते = याचनां करोति (माँगता है, याचना करता है)।

सन्दर्भ-प्रसङ्गश्च – यह पद्य हमारी शेमुषी पाठ्यपुस्तक के ‘अन्योक्तयः’ पाठ से लिया गया है। मूलतः यह पद्य पण्डितराज जगन्नाथ रचित ‘भामिनीविलास’ ग्रन्थ से लिया गया है। इस श्लोक में कवि चातक पक्षी के माध्यम से स्वाभिमान को कायम रखने के लिये कहता है।

हिन्दी-अनुवादः – एक ही स्वाभिमानी पक्षी पपीहा वन में निवास करता है (वह) या तो प्यासा (ही) मर जाता है अथवा इन्द्र से याचना करता है (किसी अन्य से नहीं)।

6. आश्वास्य पर्वतकुलं तपनोष्णतप्तमुद्दामदावविधुराणि च काननानि।
नानानदीनदशतानि च पूरयित्वा, रिक्तोऽसि यज्जलद ! सैव तवोत्तमा श्रीः।।6।।

अन्वयः – तपनोष्णतप्तम् पर्वतकुलम् आश्वास्य उद्दामदावविधुराणि काननानि च (आश्वास्य) नानानदीनदशतानि पूरयित्वा च हे जलद ! यत् रिक्तः असि तव सा एव उत्तमा श्रीः।

शब्दार्थाः – तपनोष्णतप्तम् = सूर्यस्य उष्णतया प्रतप्तम् (सूर्य की गर्मी से तपे हुए), पर्वतकुलम् = गिरीणाम् समूहम् (पर्वतों के समूह को), आश्वास्य = विश्वासम् उत्पाद्य, समाश्वास्य सन्तोष्य (सन्तुष्ट/आश्वस्त करके), उद्दामदाववि धुराणि = उन्नतकाष्ठरहितानि (ऊँचे काष्ठों अर्थात् वृक्षों से रहित), काननानि च = बनानि च (और वनों को), [आश्वास्य = समाश्वास्य (आश्वस्त करके)], नानानदीनदशतानि = अनेकसरितः नदानां शतानि च (अनेक नदियों और सैकड़ों नदों को), पूरयित्वा च = पूर्णं कृत्वा च (और पूर्ण करके, भरकर), हे जलद ! – हे वारिद ! (हे मेघ !), यत् रिक्तः असि = यत्त्वम् जलहीनः जातोऽसि (जो तुम जलहीन हो गए हो), तव सा एव = तेऽअसावेव (तेरी वही), उत्तमा श्रीः = श्रेष्ठा/श्रेष्ठतमा शोभा (अस्ति), (उत्तम शोभा है)।

सन्दर्भ-प्रसङ्गश्च- यह पद्य हमारी शेमुषी पाठ्यपुस्तक के ‘अन्योक्तयः’ पाठ से लिया गया है। मूलतः यह पद्य महाकवि माघकृत ‘शिशुपालवधम्’ महाकाव्य से संकलित है। इस श्लोक में कवि जलद (मेघ) के माध्यम से मानव को दान और परोपकार के लिये प्रेरित करता है।

हिन्दी-अनुवादः – सूर्य की गर्मी से तपे हुए पर्वतों के समूह को सन्तुष्ट करके और ऊँचे काष्ठों अर्थात् वृक्षों से रहित वनों को आश्वस्त करके अनेक नदियों और सैकड़ों नदों को भरकर हे मेघ ! जो तुम जलहीन (खाली) हो गएं हो, वही तुम्हारी उत्तम शोभा है।

7. रे रे चातक ! सावधानमनसा मित्र क्षणं श्रूयता –
मम्भोदा बहवो हि सन्ति गगने सर्वेऽपि नैतादृशाः।
केचिद् वृष्टिभिरार्द्रयन्ति वसुधां गर्जन्ति केचिद् वृथा,
यं यं पश्यसि तस्य तस्य पुरतो मा ब्रूहि दीनं वचः।।7।।

अन्वयः – रे रे मित्र चातक ! सावधानमनसा क्षणं श्रूयताम, गगने हि बहवः अम्भोदाः सन्ति, सर्वेऽपि एतादृशाः न (सन्ति), केचित् वसुधां वृष्टिभिः आर्द्रयन्ति, केचित् वृथा गजेन्ति, (त्व म्) यं यं पश्यसि तस्य तस्य पुरतः दीनं वचः मा ब्रूहि।

शब्दार्थाः – रे मित्र चातक! = हे सुहृद् सारङ्ग ! (हैं मित्र पपीहे !), सावधानमनसा = सावहितः सन् (सावधान मन से), क्षणम् = क्षणमात्रम्, किञ्चित् कालम् (क्षणभर), श्रूयताम् = आकण्यताम् श्रुणुहि (सुनिए), गगने = आकाशे (आकाश में). हि = यतः (क्योंकि). बहवः = अनेके (बहत से). अभीदाः – वारिदाः (बादल हैं). सर्वेऽपि = निः शेषाः अपि (सभी), एतादृशाः = एवं विधाः (इस प्रकार के), न सन्ति = नहि वर्तन्ते (नहीं हैं), केचित् वसुधां = केचन् पृथिवी, धरातलं (धरती को), वृष्टिभिः = वर्षया जलेन (पानी बरसाकर), आद्रयन्ति – आद्रं कुर्वन्ति, सिञ्चन्ति, क्लेदयन्ति (गीला कर देते हैं), केचित् – केचन (कुछ), वृथा = व्यर्थमेव (व्यर्थ ही), गर्जन्ति = गर्जनं कुर्वन्ति (गरजते है), [त्वम् = तुम], यं यं पश्यसि = यं यम् ईक्षसे (जिस जिसको देखते हो), तस्य तस्य पुरुतः = अमुष्य अमुष्य सम्मुखे, समक्षे (उस उसके सामने), दीनं वचः = करुणवचनं (दीन वचन), मा. न (मत), ब्रूहि = वद (बोलो)।

सन्दर्भ-प्रसङ्गश्च – यह पद्य हमारी शेमुषी पाठ्यपुस्तक के ‘अन्योक्तयः’ पाठ से लिया गया है। मूलतः यह श्लोक कवि नीतिकार भर्तृहरि रचित नीति शतक से संकलित है। इस श्लोक में कवि नीतिकार मानव को चातक के माध्यम से उपदेश देता है जिस किसी के सामने हाथ मत फैलाओ क्योंकि सभी धार्मिक (बादल) दाता नहीं होते।

हिन्दी-अनुवादः – हे मित्र पपीहे ! सावधान मन से (ध्यान से) क्षणभर सुनिए कि आकाश में बहुत से बादल हैं (परन्तु) सभी इस प्रकार के नहीं हैं। कुछ तो पानी बरसाकर धरती को गीला कर देते हैं (और) कुछ व्यर्थ ही गर्जते हैं। तुम जिस जिस को देखो उस उसके सामने दीन वचन मत बोलो अर्थात् याचना मत करो।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
1. 2x² – 3x + 5 = 0
2. 3x² – 4\(\sqrt{3}\)x + 4 = 0
3. 2x² – 6x + 3 = 0
Solution:
The given equation is of the form
ax² + bx + c = 0; where a = 2, b = -3 and c = 5.
Then, the discriminant
b² – 4ac = (-3)² – 4(2)(5)
= 9 – 40
= -31 <0
So, the given equation has no real roots.

2. Comparing the given equation with the standard quadratic equation
ax² + bx + c = 0, we get a = 3, b = -4\(\sqrt{3}\) and c = 4.
Then, the discriminant
b² – 4ac = (-4\(\sqrt{3}\))² – 4(3)(4)
= 48 – 48
= 0
So, the given equation has equal real roots.
The roots are \(-\frac{b}{2 a},-\frac{b}{2 a}\)
i.e., \(-\frac{-4 \sqrt{3}}{2(3)},-\frac{-4 \sqrt{3}}{2(3)}\), i.e., \(\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

3. The given equation is of the form
ax² + bx + c = 0, where a = 2, b = -6 and c = 3.
Then, the discriminant
b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24
= 12
So, the given equation has two distinct roots.
The roots are given by
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
\(=\frac{6 \pm \sqrt{12}}{2(2)}\)
\(=\frac{6 \pm 2 \sqrt{3}}{4}=\frac{3 \pm \sqrt{3}}{2}\)
Thus, the roots of the given equation are \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3-\sqrt{3}}{2}\)

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots:
1. 2x² + kx + 3 = 0
2. kx (x – 2) + 6 = 0
Solution:
1. Comparing the given equation with the standard quadratic equation, we have
a = 2, b = k and c = 3.
Then, the discriminant = b² – 4ac
= (k)² – 4 (2) (3)
= k² – 24
If the equation has two equal roots, then the discriminant = 0
∴ k² – 24 = 0
∴ k² = 24
∴ k = ±\(\sqrt{24}\)
∴ k ± 2\(\sqrt{6}\)

2. kx(x – 2) + 6 = 0
∴ kx² – 2kx + 6 = 0
Here, a = k, b = -2k and c = 6.
Then, the discriminant = b² – 4ac
= (-2k)² – 4(k)(6)
= 4k² – 24k
If the equation has two equal roots, then the discriminant = 0
∴ 4k² – 24k = 0
∴ 4k (k – 6) = 0
∴ k = 0 or k = 6
But k = 0 is not possible because if k = 0, the equation reduces to 6 = 0, not a quadratic equation.
∴ k = 6

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Solution:
Considering that the required mango grove can be designed, let the breadth of the mango grove be x m.
Then, the length of the mango grove is 2x m. Area of rectangular mango grove
= Length × Breadth
= 2x × x
= 2x² m²
The required area is 800 m².
∴ 2x² = 800
∴ 2x² – 800 = 0
∴ x² – 400 = 0
If the above equation has real roots, then it is possible to design the required mango grove.
Here, a = 1, b = 0 and c = -400.
Then, the discriminant = b² – 4ac
= (0)² – 4(1)(-400)
= 1600 > 0
Hence, the equation has real roots. So, it is possible to design the mango grove with required measures.
Now, x² – 400 = 0
∴ (x + 20) (x – 20) = 0
∴ x + 20 = 0 or x – 20 = 0
∴ x = -20 or x = 20
Since x is the breadth of the rectangular mango grove, x = -20 is not possible.
∴ x = 20 and 2x = 40
Thus, the length of the mango grove is 40 m and its breadth is 20 m.

Question 4.
Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the present ages of two friends be x years and (20 – x) years.
Four years ago, their respective ages were (x – 4) years and (20 – x – 4) years, i.e., (16 – x) years.
Then, according to given,
(x – 4) (16 – x) = 48
∴ 16x – x² – 64 + 4x = 48
∴ -x² + 20x – 64 – 48 = 0
∴ -x² + 20x – 112 = 0
∴ x² – 20x + 112 = 0
Here, a = 1, b = -20 and c = 112.
Then, the discriminant
b² – 4ac = (-20)² – 4(1)(112)
= 400 – 448
= -48 < 0
Hence, the equation has no real roots. So, the given situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth.
Solution:
Let the length of the rectangular park be x m.
Perimeter of rectangular park = 2 (Length + Breadth)
∴ 80 = 2(x + Breadth)
∴ 40 = x + Breadth
∴ Breadth = (40 – x) m
Now, area of rectangular park = Length × Breadth
∴ 400 = x (40 – x)
∴ 400 = 40x – x²
∴ x² – 40x + 400 = 0
Here, a = 1, b = -40 and c = 400.
Then, the discriminant = b² – 4ac
= (-40)² – 4(1)(400)
= 1600 – 1600
= 0
Hence, the equation has equal real roots. So, it is possible to design a rectangular park with given measures.
x² – 40x + 400 = 0
∴ x² – 20x – 20x + 400 = 0
∴ x(x – 20) – 20(x – 20) = 0
∴ (x – 20) (x – 20) = 0
∴ x – 20 = 0 or x – 20 = 0
∴ x = 20 or x = 20
Thus, the length of the rectangular park = x = 20 m and the breadth of the rectangular park = 40 – x = 40 – 20 = 20 m.
Note: Here the shape of the park turns out to be a square, but as we know, every square is a rectangle.