Class 9

Students should go through these JAC Class 9 Maths Notes Chapter 6 Lines and Angles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 6 Lines and Angles

Line:
A line has length but no width and no thickness.

Angle:
An angle is formed by two non-collinear rays with a common initial point. The common initial point is called the ‘vertex’ of the angle and two rays are called the ‘arms’ of the angle.

Remark:
Every angle has a measure and unit of measurement is degree. One right angle = 90°
1° = 60′ (minutes)
1′ = 60″ (seconds)

Angle addition axiom: If X is a point in the interior of BAC, then
∠BAC = ∠BAX + ∠XAC

Types of Angles:
→ Right angle: An angle whose measure is 90° is called a right angle.

→ Acute angle: An angle whose measure is more than 0° but less than 90° is called an acute angle.

0° < ∠BOA < 90°
→ Obtuse angle: An angle whose measure is more than 90° but less than 180° is called an obtuse angle.

90° < ∠AOB < 180°
→ Straight angle: An angle whose measure is 180° is called a straight angle.

→ Reflex angle: An angle whose measure is more than 180° but less than 360 is called a reflex angle.

180° < ∠AOB < 360°
→ Complementary angles: Two angles whose sum is 90° are called complementary angles.

∠AOC and ∠BOC are complementary as
∠AOC + ∠BOC = 90°
→ Supplementary angles: Two angles whose sum is 180° are called the supplementary angles.

∠AOC and ∠BOC are supplementary as their sum is 180°.
→ Angle bisectors: A ray OX is said to be the bisector of ∠AOB, if X is a point in the interior of ∠AOB, and ∠AOX = ∠BOX

→ Adjacent angles: Two angles are called adjacent angles if
→ they have the same vertex,
→ they have a common arm,
→ non-common arms are on either side of the common arm.

∠AOX and ∠BOX are adjacent angles as o is the common vertex, OX is common arm, OA and OB are non-common arms and lie on either side of OX.
→ Lincar pair of angles: Two adjacent angles are said to form a linear pair if their non-common arms are two opposite rays.

→ Vertically opposite angles: Two angles are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays.

∠AOC and ∠BOD form a pair of vertically opposite angles. Also ∠AOD and ∠BOC form a pair of vertically opposite angles.

Angles Made by a Transversal with two Parallel Lines:
→ Transversal: A line which intersects two or more given lines at distinct points is called a transversal of the given lines

→ Corresponding angles: Two angles on the same side of transversal are known as the corresponding angles if both lie either above the two lines or below the two lines, in figure 1 and 5, 4 and 8, 2 and 6, 3 and 7 are the pairs of corresponding angles.
→ Alternate interior angles: ∠3 and ∠5, ∠2 and ∠8, are the pairs of alternate interior angles.
→ Consecutive interior angles: The pair of interior angles on the same side of the transversal are called pair of consecutive interior angles. In figure ∠2 and ∠5, ∠3 and ∠8. are the pairs of consecutive interior angles.
→ Corresponding angles axiom: If a transversal intersects two parallel lines, then each pair of corresponding angles are equal. Conversely, if a transversal intersects two lines, making a pair of equal corresponding angles, then the lines are parallel.

Important Facts to Remember:
→ If a ray stands on line, then the sum of the adjacent angles so formed is 180°.
→ If the sum of two adjacent angles is 180°, then their non-common arms are two apposite rays.
→ The sum of all the angles round a point is equal to 360°.
→ If two lines intersect, then the vertically opposite angles are equal.
→ If a transversal interests two parallel lines then the corresponding angles are equal, each pair of alternate interior angles is equal and each pair of consecutive interior angles is supplementary.
→ If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are parallel.
→ If a transversal intersects two lines in such a way that a pair of consecutive interior angles are supplementary, then the two lines are parallel.
→ If two parallel lines are intersected by a transversal, the bisectors of any pair of alternate interior angles are parallel and the bisectors of any two corresponding angles are also parallel.
→ If a line is perpendicular to one of the two given parallel lines, then it is also perpendicular to the other line.
→ Two angles which have their arms parallel are either equal or supplementary.
→ Two angles whose arms are perpendicular are either equal or supplementary.

Important Theorems:
Theorem 1.
If two lines intersect each other, then the vertically opposite angles are equal.
Proof:
Given: Two lines AB and CD intersecting at a point O.

To prove: ∠AOC = ∠BOD, ∠BOC
= ZAOD
Proof: Since ray OD stands on AB
again, ray OA stands on CD
⇒ ∠AOD + ∠DOB = 180° [linear pair] ……(i)
⇒ ∠AOC + ∠AOD = 180° [linear pair] ……(ii)
From equations (i) and (ii), we get
∠AOD + ∠DOB
= ∠AOC + ∠AOD
⇒ ∠AOC = ∠DOB
Similarly, we can prove that
∠BOC = ∠DOA Hence, proved.

Theorem 2.
If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Proof:
Given: AB and CD are two parallel lines. Transversal l intersects AB and CD at P and Q respectively making two pairs of alternate interior angles, ∠1, ∠2 and ∠3, ∠4.

To prove: ∠1 = ∠2 and ∠3 = ∠4
Proof: Clearly, ∠2 = ∠5 [Vertically opposite angles]
And, ∠1 = ∠5 [Corresponding angles]
∴ ∠1 = ∠2
Also, ∠3 = ∠6 [Vertically opposite angles]
And, ∠4 = ∠6 [Corresponding angles]
∴ ∠3 = ∠4 Hence, proved.

Jharkhand Board JAC Class 9 Maths Important Questions Chapter 8 Quadrilaterals Important Questions and Answers.

JAC Board Class 9th Maths Important Questions Chapter 8 Quadrilaterals

Question 1.
ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and BC respectively. If AB = 12 cm and MN = 14 cm, find CD.

Solution :
Here, ABCD is a trapezium in which, AB || DC and M and N are mid-points of AD and BC respectively. Since the line segment joining the midpoints of non-parallel sides of trapezium is half of the sum of the lengths of its parallel sides
⇒ MN = \(\frac {1}{2}\)(AB + CD)
⇒ 14 = \(\frac {1}{2}\)(12 + CD)
⇒ 28 = 12 + CD
⇒ CD = 28 – 12 = 16 cm

Question 2.
Use the informations given in figure below to calculate the value of x.

Solution :
Since, EB is a straight line.
∴ ∠DAE + ∠DAB = 180°
⇒ 73° + ∠DAB = 180°
i.e., ∠DAB = 180° – 73° = 107°
∴ ∠DAB + ∠ABC + ∠BCD + ∠CDA = 360°
Since the sum of the angles of quadrilateral ABCD is 360°
∴ 107° + 105° + x + 80° = 360°
⇒ 292° + x = 360°
⇒ x = 360° – 292° = 68°

Question 3.
ABCD is a rhombus and AB is produced to E and F such that AE = AB = BF. Prove that EG and FG are perpendicular to each other.

Solution :
Given: ABCD is a rhombus. AB produced to E and F such that AE = AB = BF
Construction: Join ED and CF and produce it to meet at G.
To prove: ED ⊥ FC
Proof: AB is produced to points E and F such that AE = AB = BF …(i)
Also, since ABCD is a rhombus
AB = CD = BC = AD ……..(ii)
Now, in ΔBCF, BC = BF [From (i) and (ii)]
⇒ ∠1 = ∠2
∠3 = ∠1 + ∠2 [Exterior angle]
∠3 = 2∠2 ……..(iii)
Similarly, AE = AD
∠5 = ∠6 …(iv)
⇒ ∠4 = ∠5 + ∠6 = 2∠5
Adding (iii) and (iv) we get
∠4 + ∠3 = 2∠5 +2∠2
⇒ 180° = 2(∠5 + ∠2) [∵ ∠4 and ∠3 are consecutive interior angles]
⇒ ∠5 + ∠2 = 90°
∴ Now in ΔEGF
∠5 + ∠2 + ∠EGF = 180°
⇒ ED ⊥ FC
⇒ ∠EGF = 90° Hence Proved.

Question 4.
In the given figure, E and F are respectively, the mid-points of nonparallel sides of a trapezium ABCD. Prove that
(i) EF || AB
(ii) EF = \(\frac {1}{2}\)(AB + DC).

Solution :
Join BE and produce it to intersect CD produced at point P. In ΔAEB and ΔDEP, AB || PC and BP is transversal
⇒ ∠ABE – ∠DPE (Alternate interior angles)
∠AEB = ∠DEP (Vertically opposite angles)
And AE = DE (E is mid-point of AD)
⇒ ΔAEB ≅ ΔDEP (By AAS)
⇒ BE = PE [By CPCT]
And AB = DP [By CPCT]
Since the line segment joining the midpoints of any two sides of a triangle is parallel and half of the third side, therefore, in ΔBPC E is mid-point of BP [As, BE = PE]
and F is mid-point of BC [Given]
⇒ EF || PC and EF = \(\frac {1}{2}\)PC
⇒ EF || DC and EF = \(\frac {1}{2}\)(PD + DC)
⇒ EF || AB and EF = \(\frac {1}{2}\)(AB + DC) (As, DC || AB and PD = AB)
Hence, proved.

Multiple Choice Questions

Question 1.
In a parallelogram ABCD, ∠D = 105°, then the ∠A and ∠B will be :
(a) 105°, 75°
(b) 75°, 105°
(c) 105°, 105°
(d) 75°, 75°
Solution :
(b) 75°, 105°

Question 2.
In a parallelogram, ABCD diagonals AC and BD intersect at O and AC = 12.8 cm and BD = 7.6 cm, then the measures of OC and OD respectively equal to :
(a) 1.9 cm, 6.4 cm
(b) 3.8 cm, 3.2 cm
(c) 3.8 cm, 3.2 cm
(d) 6.4 cm, 3.8 cm
Solution :
(d) 6.4 cm, 3.8 cm

Question 3.
Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)° then the value of x will be :
(a) 17°
(b) 16°
(c) 15°
(d) 13°
Solution :
(d) 13°

Question 4.
When the diagonals of a parallelogram are perpendicular to each other then it is called a :
(a) Square
(b) Rectangle
(c) Rhombus
(d) Trapezium
Solution :
(c) Rhombus

Question 5.
In a parallelogram ABCD, E is the midpoint of side BC. If DE and AB when produced meet at F then: (a) AF = \(\frac {1}{2}\)AB
(b) AF = 2AB
(c) AF = 4AB
(d) Data Insufficient
Solution :
(b) AF = 2AB

Question 6.
ABCD is a rhombus with ∠ABC = 56°, then the ∠ACD will be:
(a) 56°
(b) 62°
(c) 124°
(d) 34°
Solution :
(b) 62°

Question 7.
In a triangle, P, Q and R are the midpoints of the sides BC, CA and AB respectively. If AC = 16 cm, BC = 20 cm and AB = 24 cm then the perimeter of the quadrilateral ARPQ will be:
(a) 60 cm
(b) 30 cm
(c) 40 cm
(d) None of these
Solution :
(c) 40 cm

Question 8.
LMNO is a trapezium with LM || NO. If P and Q are the mid-points of LO and MN respectively and LM = 5 cm and ON = 10 cm then PQ =
(a) 2.5 m
(b) 5 cm
(c) 7.5 cm
(d) 15 cm
Solution :
(c) 7.5 cm

Question 9.
In an isosceles trapezium ABCD if ∠A = 45° then ∠C will be:
(a) 90°
(b) 135°
(c) 125°
(d) None of these
Solution :
(b) 135°

Question 10.
In a right-angle triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, then the area of ΔADE =
(a) 67.5 cm²
(b) 13.5 cm²
(c) 27 cm²
(d) Data insufficient
Solution :
(b) 13.5 cm²

Question 11.
When the opposite sides of quadrilateral are parallel to each other then it is called a:
(a) Square
(b) Parallelogram
(c) Trapezium
(d) Rhombus
Solution :
(b) Parallelogram

Question 12.
In the given figure, AP and BP are angle bisectors of ∠A and ∠B which meet at P in the parallelogram ABCD. Then 2∠APB =

(a) ∠A + ∠B
(b) ∠A + ∠C
(c) ∠B + ∠D
(d) 2∠C + ∠B
Solution :
(a) ∠A + ∠B

Question 13.
In a quadrilateral ABCD, AO and DO are angle bisectors of ∠A and ∠D and given that ∠C = 105°, ∠B = 70° then the ∠AOD is :
(a) 67.5°
(b) 77.5°
(c) 87.5°
(d) 99.75°
Solution :
(c) 87.5°

Question 14.
In a parallelogram the sum of the angle bisectors of two adjacent angles is :
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution :
(d) 90°

Jharkhand Board JAC Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

Page-164

Question 1.
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Answer:
In ∆AFD,
∠F = 90° [Each angle of a rectangle is equal to 90°]
∠F + ∠A + ∠D = 180°
(Angle sum property)
⇒ 90° + ∠A + ∠D = 180°
⇒ ∠A + ∠D = 180° – 90° = 90°
⇒ ∠D < 90° ( v ∠F = 90°)
⇒ ∠D < ∠F
⇒ AF < AD [Since side opposite to larger angle is longer] Adding AB to both the sides AD + AB > AF + AB
Multiplying by 2
2 [AD + AB] > 2 [AF + AB]
⇒ Perimeter of the parallelogram ABCD > Perimeter of the Rectangle ABEF.

Question 2.
In Figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).

Answer:
In ∆ABE, AD is median [∵ BD = DE]
So, ar (ABD) = ar (AED) ………..(i)
[∵ Median of a triangle divides it into two parts of equal areas.]
Similarly,
In ∆ADC, AE is median [∵ DE = EC]
So, ar (ADE) = ar (AEC) ………..(ii)
From equations (i) and (ii), we get
ar (ABD) = ar (ADE) = ar (AEC)

Page-165

Question 3.
In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

Answer:
In ∆ADE and ∆BCF,
AD = BC [∵ Opposite sides of parallelogram ABCD]
DE = CF [∵ Opposite sides of parallelogram DCFE]
AE = BF [∵ Opposite sides of parallelogram ABFE]
So, ∆ADE ≅ ∆BCF [∵ SSS Congruence theorem]
∴ ar (ADE) = ar (BCF) [∵ Congruent triangles have equal areas]

Question 4.
In Fig, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).

Answer:
In ∆ADP and ∆QCP,
∠APD = ∠QPC [∵ Vertically Opposite Angles]
∠ADP = ∠QCP [∵ Alternate angles]
AD = CQ [∵ Given]
So, ∆APD ≅ ∆CPQ [∵ AAS Congruence theorem]
So, DP = CP [∵ CPCT]
In ∆CDQ, QP is median. [∵ DP = CP]
So, ar (DPQ) = ar (QPC) …(i)
[∵ Median of a triangle divides it into two parts of equal areas.]
Similarly,
In ∆PBQ, PC is median.
[∵ AD = CQ and AD = BC ⇒ BC = QC]
So, ar (QPC) = ar (BPC) …(ii)
From equations (i) and (ii), we get
ar (BPC) = ar (DPQ)

Question 5.
In Figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(i) ar (BDE) = \(\frac{1}{4}\) ar (ABC)
(ii) ar (BDE) = \(\frac{1}{2}\) ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = \(\frac{1}{8}\) ar (AFC)
[Hint: Join EC and AD. Show that BE 11 AC and DE|| AB, etc.]
Answer:
(i) Construction: Join EC and AD

Let, BC = x
So, ar(∆ABC) = \(\frac{\sqrt{3}}{4}\) x²
[∵ Area of a equilateral triangle = \(\frac{\sqrt{3}}{4}\) (Side)²
and ar (∆BDE) = \(\frac{\sqrt{3}}{4}\left(\frac{x}{2}\right)^2\)
[∵ D is the mid-point of BC]
= \(\frac{1}{4}\) [\(\frac{\sqrt{3}}{4}\) x²] = \(\frac{1}{4}\) [ar(AABC)]

(ii) In ∆BEC, ED is median
[∵ D is the mid-point of BC]
So, ar (BDE) = \(\frac{1}{2}\) ar (∆BEC) …(i)
[∵ Median of a triangle divides it into two parts of equal areas.]
∠EBC = 60° and ∠BCA = 60°
[∵ Angles of equilateral triangle]
So, ∠AEBC = ∠BCA
Since, alternate angles (∠EBC = ∠BCA) are equal, so BE || AC
Triangles BEC and BAE are on the same base BE and between same parallels BE || AC
So, ar (∆BEC) = ar (∆BAE) …(ii)
[∵ Triangles on the same base and between same parallels are equal in area]
From equation (i) and (ii), we get
So, ar (∆BDE) = \(\frac{1}{2}\) ar (∆BAE)

(iii) In ∆BEC, ED is median
[∵ D is the mid-point of BC]
So, ar (∆BDE) = \(\frac{1}{2}\) ar (∆BEC) …(iii)
[∵ Median of a triangle divides it into two parts of equal areas.]
ar (∆BDE) = \(\frac{1}{2}\) ar (∆ABC) ………..(iv)
[∵ Proved in (i)]
From the equations (iii) and (iv), we get
ar (∆ABC)= 4 ar (∆BDE) = 4 (\(\frac{1}{2}\)) ar(∆BEC) = 2 ar (∆BEC)

(iv) ∠ABD = 60° and ∠BDE = 60°
[∵ Angles of equilateral triangle]
So, ∠ABD = ∠BDE
Since, alternate angles (∠ABD = ∠BDE) are equal, so BA || ED Triangles BDE and AED are on the same base ED and between same parallels BA || ED.
So, ar (∆BDE) = ar (∆AED)
[∵ Triangles on the same base and betwee same parallels are equal in area]
Subtracting ar (∆FED) from both the sides
ar (∆BDE) – ar (∆FED) = ar (∆AED) – ar (∆FED)
⇒ ar (∆BEF) = ar (∆AFD)

(v) ΔADF is also right angled at D. [As in equilateral triangle, median and altitude are same]
⇒ AB² = AD² + BD²
⇒ AD² = AB² – BD²
Let AB = a and hence BD = \(\frac{a}{2}\)
(as AD is the median)
⇒ AD = \(\frac{\sqrt{3}a}{2}\)
In ΔFED,
EF² = DE² – DF² = (\(\frac{a}{2}\))² – (\(\frac{a}{4}\))²
= \(\frac{a^{2}}{4}\) – \(\frac{a^{2}}{10}\) = \(\frac{3a^{2}}{16}\)
⇒ EF = \(\frac{\sqrt{3}a}{2}\)

(vi) ar (ΔBDE) = \(\frac{1}{4}\) ar (ΔABC) [∵ From Part (i)]
⇒ ar (ΔBEF) + ar (ΔFED) = \(\frac{1}{4}\) ar (ΔABC)
⇒ ar (ΔBEF) + ar (ΔFED) = \(\frac{1}{4}\) [2 ar (ΔADC)]
[∵ ar (ΔABC) = 2 ar (ΔADC)]
⇒ 2 ar (ΔFED) + ar (ΔFED) = \(\frac{1}{2}\) ar (ΔADC) [∵ From Part (v)]
⇒ 3 ar (ΔFED) = \(\frac{1}{2}\) [ar (ΔAFC) – ar (ΔAFD)]
⇒ 3 ar (ΔFED) = \(\frac{1}{2}\) [ar (ΔAFC) – 2ar (ΔFED)] [∵ From Part (vii)]
⇒ 3 ar (ΔFED) = \(\frac{1}{2}\) ar (ΔAFC) – \(\frac{1}{2}\) × 2ar (ΔFED)
⇒ 3 ar (ΔFED) = \(\frac{1}{2}\) ar (ΔAFC) – ar (ΔFED)
⇒ 4 ar (ΔFED) = \(\frac{1}{2}\) ar (ΔAFC)
⇒ ar (ΔFED) = \(\frac{1}{8}\) ar (ΔAFC)

Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint: From A and C, draw perpendiculars to BD]
Answer:
Construction: From A and C, draw perpendiculars AM and CN to BD.
ar (ΔAPB) × ar (ΔCPD) = \(\frac{1}{2}\) × BP × AM × \(\frac{1}{2}\) × PD × CN …(i)

ar (ΔAPD) × ar (ΔBPC) = \(\frac{1}{2}\) × PD × AM × \(\frac{1}{2}\) × BP × CN …(ii)
From (i) and (ii), we get
ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC)

Question 7.
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar (PRQ) = \(\frac{1}{2}\) ar (ARC)
(ii) ar (RQC) = \(\frac{3}{8}\) ar (ABC)
(iii) ar (PBQ) = ar (ARC)
Answer:
Construction: Join AQ, PC, RC and RQ.

(i) In ΔAPQ, QR is median [∵ Given]
So, ar (ΔPQR) = \(\frac{1}{2}\) ar (ΔAPQ) …(i)
[∵ Median of a triangle divides it into two parts of equal areas.]
Similarly,
In ΔAQB, QP is median [∵ Given]
[∵ So, ar (ΔAPQ) = \(\frac{1}{2}\) ar (ΔABQ) ……(ii)
and in ΔABC, AQ is median [∵ Given]
So, ar (ΔABQ) = \(\frac{1}{2}\) ar (ΔABC) ……..(iii)
Form (i), (ii) and (iii), we get
ar (ΔPQR) = \(\frac{1}{2}\) ar (ΔAPQ)
= \(\frac{1}{2}\)[\(\frac{1}{2}\) ar(ΔABQ)]
= \(\frac{1}{4}\)[\(\frac{1}{2}\)ar(ΔABC)]
= \(\frac{1}{8}\) ar (ABC) ………(iv)
In ΔAPC, CR is median. [∵ Given]
So, ar (ΔARC) = \(\frac{1}{2}\) ar (APC) ……(v)
[∵ Median of a triangle divides it into two parts of equal areas.]
Similarly,
In ΔABC, CP is median [∵ Given]
So, ar (ΔAPC) = \(\frac{1}{2}\) ar (ΔABC) ……..(vi)
Form (v) and (vi), we get
ar (ΔARC) = \(\frac{1}{2}\) ar (ΔAPC)
= \(\frac{1}{2}\)[\(\frac{1}{2}\) ar(ΔABC)]
= \(\frac{1}{2}\)[\(\frac{1}{2}\) ar(ΔABC)]
= \(\frac{1}{4}\) ar(ΔABC) ……..(vii)
Form (iv) and (vii), we get
ar (ΔPQR) = \(\frac{1}{8}\) ar (ΔABC)
= \(\frac{1}{2}\)[\(\frac{1}{4}\) ar(ΔABC)]
= \(\frac{1}{2}\)[\(\frac{1}{2}\) ar(ΔABC)]
= \(\frac{1}{8}\) ar(ΔARC)

(ii) ar (∆RQC) = = ar (∆RQA) + ar (∆AQC) – ar (∆ARC) ……(viii)
In ∆PQA, QR is median [∵ Given]
So, ar (RQA) = \(\frac{1}{2}\) ar (PQA) …(ix)
In ∆AQB, PQ is median. [∵ Given]
So, ar (PQA) = \(\frac{1}{2}\) ar (AQB) …….(x)
In ABC, AQ is median [∵ Given]
So, ar (∆AQB) = \(\frac{1}{2}\) ar (∆ABC) …(xi)
From (ix), (x) and (xi), we get

(iii) In ΔABQ, PQ is median [∵ Given]
So, ar (ΔPBQ) = \(\frac{1}{2}\) ar (ΔABQ) ……..(xvii)
In ΔABC, AQ is median
So, ar (ΔABQ) = \(\frac{1}{2}\) ar (ΔABC) ……..(xviii)
Form (xvi), (xvii) and (xviii), we get
ar (ΔARC) = \(\frac{1}{4}\) ar (ΔABC)
= \(\frac{1}{4}\)[2 ar(ΔABQ)]
= \(\frac{1}{2}\) (2) ar(ΔPBQ)
= ar(ΔPBQ)

Question 8.
In the figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:

(i) ∆MBC ≅ ∆ABD
(ii) ar (BYXD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar (CYXE) = 2 ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
Answer:
(i) In ∆MBC and ∆ABD,
BC = BD [ v Sides of square]
∠MBC = ∠ABD = 90° + ∠ABC
MB = AB [∵ Sides of square]
So, ∆MBC ≅ ∆ABD
[∵ SAS Congurence theorem]

(ii) YX || BD
[Opposite sides of square are parallel]
Triangle ABD and parallelogram BYXD are on the same base BD and lie between the same parallels AX j | BD.
So, ar (AABD) = \(\frac{1}{2}\) ar (BYXD) …(i)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.]
But, ∆MBC ≅ ∆ABD
[∵ Proved above]
So, ar (MBC) = ar (ABD) …(ii)
From (i) and (ii), we get
ar (∆MBC) = ar (∆ABD)
= \(\frac{1}{2}\) ar (BYXD) …(iii)
⇒ 2 ar (MBC) = ar (BYXD)

(iii) MB || NA (Opposite sides of square are parallel)
Triangle MBC and square ABMN are on the same base MB and lie between the same parallels MB || NC.
So, ar (∆MBC) = \(\frac{1}{2}\) ar (ABMN)…(iv)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.]
From (iii) and (iv), we get
ar (BYXD) = 2 ar (MBC) = ar (ABMN)

(iv) In ∆ACE and ∆BCF,
CE = BC [∵ Sides of square]
∠ACE = ∠BCF = 90° + ∠ACB
AC = CF [∵ Sides of square]
So, ΔACE ≅ ΔBCF
[∵ SAS Congruence rule]

(v) EC || XY (Opposite sides of square are parallel)
Triangle ACE and square CYXE are on the same base CE and lie between same parallels CE || AX.
So, ar (ACE) = \(\frac{1}{2}\) ar (CYXE)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.]
⇒ ar (AFCB) = \(\frac{1}{2}\) ar (CYXE) …(v)
[As ΔACE ≅ ΔBCF ⇒ ar (FCB) = ar (AACE)]
⇒ 2 ar (AFCB) = ar (CYXE)

(vi) CF || AG (Opposite sides of square are parallel)
Triangle BCF and square ACFG are on the same base CF and lie between same parallels CF || FG.
So, ar (ABCF) = \(\frac{1}{2}\) ar (ACFG) …(vi)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.]
From (v) and (vi), we get
⇒ ac (CYXE) = 2ar (FCB)
= 2 (\(\frac{1}{2}\))ar (ACFG) = ar (ACFG)

(vii) From part (iii) and (vi), we get
ar (BYXD) = ar (ABMN)
and ar (CYXE) = ar (ACFG)
Adding both, we get
ar (BYXD) + ar (CYXE) = ar (ABMN) + ar (ACFG)
⇒ ar (BCED) = ar (ABMN)+ ar (ACFG)

Students should go through these JAC Class 9 Maths Notes Chapter 10 Circles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 10 Circles

Definitions
→ Circle:

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle.
In figure, O is the centre and the length OP is the radius of the circle. So the line segment joining the centre and any point on the circle is called a radius of the circle.

→ Interior and Exterior of a Circle:

A circle divides the plane on which it lies into three parts. They are
(i) inside the circle (or interior of the circle)
(ii) the circle and
(iii) outside the circle (or exterior of the circle.)
The circle and its interior make up the circular region.

→ Chord:
If we take two points P and Q on a circle, then the line segment PQ is called a chord of the circle.

→ Diameter:

The chord which passes through the centre of the circle, is called a diameter of the circle.
A diameter is the longest chord and all diameters have the same length, which is equal to two times the radius. In figure, AOB is a diameter of circle.

→ Arc:
A piece of a circle between two points is called an arc. If we look at the pieces of the circle between two points Pand Q in figure, we find that there are two pieces, one longer and the other smaller. The longer one is called the major arc PQ and the shorter one is called the minor arc PQ. The minor arc PQ is also denoted by PQ and the major arc PQ by PRO, where is some point on the are between P and Q. Unless otherwise stated, arc PQ or PQ stands for minor arc PQ. When P and Q are ends of a diameter, then both arcs are equal and each is called a semicircle.

→ Circumference: The length of the complete circle is called its circumference.

→ Segment:

The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle. There are two types of segments which are the major segment and the minor segment (as in figure).

→ Sector:
The region between an arc and the two radil, joining the centre to the end points of the arc is called a sector. Like segments, we find that the minor arc corresponds to the minor sector and the major arc corresponds to the major sector. In figure, the region OPQ is the minor sector and the remaining part of the circular region is the major sector. When two arcs are equal, then both segments and both sectors become the same and each is known as a semicircular region.

Theorem 1.
Equal chords of a circle subtend equal angles at the centre.

Given: AB and CD are the two equal chords of a circle with centre O.
To Prove: ∠AOB = ∠COD.
Proof:
In ΔAOB and ΔCOD,
OA = OC [Radii of a circle]
OB = OD [Radii of a circle]
AB = CD [Given]
∴ ΔAOB ≅ ΔCOD [By SSS]
∴ ∠AOB = ∠COD. [By CPCT]

Converse of above Theorem:
If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Given: ∠AOB and ∠POQ are two equal angles subtended by chords AB and PQ of a circle at its centre O.
To Prove: AB = PQ
Proof:
In ΔAOB and ΔPOQ.
OA = OP [Radii of a circle]
OB = OQ [Radii of a circle]
∠AOB = ∠POQ [Given]
∴ ΔAOB ≅ ΔPOQ [By SAS]
∴ AB = PQ [By CPCT]
Hence, proved.

Theorem 2.
The perpendicular from the centre of a circle to a chord bisects the chord.

Given: A circle with centre O. AB is a chord of this circle OM ⊥ AB.
To Prove: MA = MB.
Construction: Join OA and OB.
Proof:
In right triangles OMA and OMB.
OA = OB [Radii of a circle]
OM = OM [Common]
∠OMA = ∠OMB [90° each]
∴ ΔOMA ≅ ΔOMB [By RHS]
∴ MA = BR [By CPCT]
Hence, proved.

Converse of above Theorem:
The line drawn through the centre of a circle to bisect a chord is a perpendicular to the chord.

Given: A circle with centre O. AB is a chord of this circle whose mid-point is M.
To Prove: OM ⊥ AB
Construction: Join OA and OB.
Proof:
In ΔOMA and ΔOMB,
MA = MB [Given]
OM = OM [Common]3
ОА = ОВ [Radii of a circle]
∴ ΔOMA ≅ ΔOMB [By SSS]
∴ ∠AMO = ∠BMO [By CPCT]
But, ∠AMO + ∠BMO = 180° [Linear pair axiom]
∴ ∠AMO = ∠BMO = 90°
⇒ OM ⊥ AB.

Theorem 3.
There is one and only one circle passing through three given non-collinear points.

Proof:
Let us take three points A, B and C, which are not on the same line, or in other words, they are not collinear [as in figure]. Draw perpendicular bisectors of AB and BC say, PQ and RS respectively. Let these perpendicular bisectors intersect at one point O. (Note that PQ and RS will intersect because they are not parallel) (as in figure).
∴ Olies on the perpendicular bisector PQ of AB.
∴ OA = OB
[ Every point on the perpendicular bisector of a line segment is equidistant from its end points]
∴ Similarly : Olies on the perpendicular bisector RS of BC.
∴ OB = OC
[∵ Every point on the perpendicular bisector of a line segment is equidistant from its end points]
So, OA = OB = OC
i.e., the points A, B and C are at equal distances from the point O.

So, if we draw a circle with centre O and radius OA it will also pass through B and C. This shows that there is a circle passing through the three points A, B and C. We know that two lines (perpendicular bisectors) can intersect at only one point. so we can draw only one circle with radius OA. In other words, there is a unique circle passing through A, B and C. Hence Proved.

Remark:
If ABC is a triangle, then by above theorem, there is a unique circle passing through the three vertices A, B and C of the triangle. This circle is the circumcircle of the ΔABC. Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.

Theorem 4.
Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Given: A circle has two equalchords AB and CD.
So, AB = CD and OM ⊥ AB, ON ⊥ CD.
To Prove: OM = ON
Construction: Join OB and OD.
Proof:
AB = CD (Given)
∴ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)CD
⇒ BM = DN
[∵ The perpendicular drawn from the centre of a circle to the chord bisect the chord.]
In ΔΟΜB and ΔOND,
∠OMB = ∠OND = 90° [Given]
OB = OD [Radii of same circle]
BM = DN [Proved above]
∴ ΔOMB ≅ ΔOND [By RHS]
OM = ON [By CPCT]
Hence, proved.

Remark: Chords equidistant from the centre of a circle are equal in length.

Some Important Theorems

Theorem 1.
If the angles subtended by the chords at the centre of a circleare equal then the chords are equal.

Given: A circle with centre O. Chords PQ and RS subtend equal angles at the center of the circle.
i.e. ∠POQ = ∠ROS
To Prove: Chord PQ = chord RS.
Proof:
In ΔPOQ and ΔROS
∠POQ = ∠ROS [Given]
OP = OR [Radii of the same circle]
OQ = OS [Radii of the same circle]
⇒ ΔPOQ ≅ ΔROS [By SAS]
⇒ chord PQ = chord RS [By CPCT]
Hence, proved.

Corollary 1: Two ares of a circle are congruent, if the angles subtended by them at the centre are equal.

Corollary 2: If two arcs of a circle are equal, they subtend equal angles at the centre.
Corollary 3: If two arcs of a circle are congruent (equal), their corresponding chords are also equal. Conversely, if two chords of a circle are equal, their corresponding arcs are also equal
∠AOB = ∠COD
∴ Chord AB = Chord CD
∴ Arc APB = Arc CQD.

Theorem 2.
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Given: An arc PQ of a circle subtending angles POQ at the centre and PAQ at a point A on the remaining part of the circle.
To Prove: ∠POQ = 2∠PAQ.

Construction: Join AO and extend it to a point B.
Proof:
There arises three cases:
(A) arc PQ is minor
(B) arc PQ is a semi-circle
(C) arc PQ is major.
In all the cases,
∠BOQ = ∠OAQ + ∠AQO ….(i)
[∵ An exterior angle of triangle is equal to the sum of the two interior opposite angles]
In OAO,
OA = OQ [Radil of a circle]
∠OAQ = ∠OQA …..(ii)
[Angles opposite to equal sides of a triangle are equal]
From (i) and (ii),
∠BOQ = 2∠OAQ ……(iii)
Similarly,
∠BOP = 2∠OAP ……(iv)
Adding (i) and (iv), we get
∠BOP + ∠BOQ = 2(∠OAP + ∠OAQ)
⇒ ∠POQ = 2∠PAQ ….(v)
NOTE: For the case (C), where PQ is the major arc, angle is replaced by reflex angles
Thus, ∠POQ = 2∠PAQ.

Theorem 3.
Angles in the same segment of a circle are equal.

Proof:
Let P and Q be any two points on a circle to form a chord PQ, A and C any other points on the remaining part of the circle and O be the centre of the circle. Then,
∠POQ = 2∠PAQ ……(i)
And ∠POQ = 2∠PCQ ……(ii)
From above equations, we get
2∠PAQ = 2∠PCQ
⇒ ∠PAQ = ∠PCQ Hence, proved

Theorem 4.
Angle in the semicircle is a right angle.

Proof:
∠PAQ is an angle in the segment, which is a semicircle.
∴ ∠PAQ = \(\frac{1}{2}\)∠POQ = \(\frac{1}{2}\) × 180° = 90°
[∵ ∠POQ is straight line angle or ∠POQ = 180°]
If we take any other point C on the semicircle then again we get
∠PCQ = \(\frac{1}{2}\)∠POQ = \(\frac{1}{2}\) × 180° = 90°
Hence, proved.

Theorem 5.
If a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment then the four points lie on a circle (i.e., they are concyclic).
Given: AB is a line segment, which subtends equal angles at two points C and D i.e, ∠ACB = ∠ADB.
To Prove: The points A, B, C and D lie on a circle.

Proof:
Let us draw a circle through the points A, C and B. Suppose it does not pass through the point D.
Then it will intersect AD (or extended AD) at a point, say E (or E’). If points A, C, E and B lie on a circle.
∠ACB = ∠AEB [∵ Angles in the same segment of circle are equal]
But it is given that, ∠ACB = ∠ADB
Therefore, ∠AEB = ∠ADB
This is possibleonly when Ecoincides with D.
[As otherwise ∠AEB > ∠ADB]
Similarly, E’ should also coincide with D.
So A, B, C and D are concyclic.
Hence, proved.

Cyclic Quadrilateral
A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle.

Theorem 6.
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

Given: A cyclic quadrilateral ABCD.
To Prove: ∠A + ∠C = ∠B + ∠D = 180°
Construction: Join AC and BD.

Proof:
∠ACB = ∠ADB [Angles in the same segment]
And ∠BAC = ∠BDC [Angles in the same segment]
∴ ∠ACB + ∠BAC = ∠ADB + ∠BDC
= ∠ADC.
Adding ∠ABC to both sides, we get
∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC
The left side being the sum of three angles of AABC is equal to 180°.
∴ ∠ADC + ∠ABC = 180°
i.e, ∠D + ∠B = 180°
∴ ∠A + ∠C = 360° – (∠B + ∠D) = 180°
[∵ ∠A + ∠B + ∠C + ∠D = 360°]
Hence, proved.
Corollary: If the sum of a pair of opposite angles of a quadrilateral is 180°, then quadrilateral is cyclic.

Jharkhand Board JAC Class 9 Sanskrit Solutions Shemushi Chapter 1 भारतीवसन्तगीतिः Textbook Exercise Questions and Answers.

JAC Board Class 9th Sanskrit Solutions Shemushi Chapter 1 भारतीवसन्तगीतिः

JAC Class 9th Sanskrit भारतीवसन्तगीतिः Textbook Questions and Answers

1. एकपदेन उत्तरं लिखत-(एक शब्द में उत्तर दीजिये-)
(क) कविः कां सम्बोधयति? (कवि किसको संबोधित करता है?)
उत्तरम् :
वाणीम् (वाणी को)।

(ख) कविः वाणीं कां वादयितुं प्रार्थयति? (कवि वाणी से क्या बजाने की प्रार्थना करता है?)
उत्तरम् :
वीणाम्। (वीणा को)

(ग) कीदृशीं वीणां निनादयितुं प्रार्थयति? (कैसी वीणा बजाने की प्रार्थना करता है?)
उत्तरम् :
नवीनाम् (नई वीणा को)।

(घ) गीति कथं गातुं कथयति? (कैसा गीत गाने के लिए कहता है?)
उत्तरम् :
नीतिलीनाम् (नीति से पूर्ण)।

(ङ) सरसाः रसालाः कदा लसन्ति? (रसीले आम कब शोभा देते हैं ?)
उत्तरम् :
वसन्ते (वसन्तु ऋतु में)।

2. पूर्णवाक्येन उत्तरं लिखत-(पूरे वाक्य में उत्तर दीजिये-)
(क) कविः वाणी किं कथयति? (कवि वाणी को क्या कहता है?)
उत्तरम् :
कविः वाणी नवीनां वीणां निनादयितुं कथयति। (कवि वाणी को नई वीणा बजाने के लिए कहता है।)

(ख) वसन्ते किं भवति? (वसन्त ऋतु में क्या होता है?)
उत्तरम् :
वसन्ते मधुर-मञ्जरी-भूतमालाः सरसा:-रसालाः लसन्ति। (वसन्त में मधुर मंजरियों से पीले वर्ण से युक्त रसीले आमों के वृक्ष शोभा देते हैं।)

(ग) सलिलं तव वीणामाकर्ण्य कथम् उच्चलेत्? (पानी तुम्हारी वीणा को सुनकर कैसे उछलता है?)
उत्तरम् :
तव वीणामाकर्ण्य सलीलं जलमुच्चलेत्। (तुम्हारी वीणा को सुनकर पानी खेल ही खेल में उछलता है।)

(घ) कविः भगवतीं भारती कस्याः तीरे मधुमाधवीनां नतां पंक्तिम् अवलोक्य वीणां वादयितुं कथयति? (कवि भगवती भारती (सरस्वती) से किस नदी के तट पर (कहाँ) मधुमाधवी की झुकी हुई पंक्तियों को देखकर वीणा को बजाने के लिए कहता है?)
उत्तरम् :
कविः भगवर्ती भारती कलिन्दात्मजायाः सवानीरतीरे मधुमाधवीनां नतां पंक्तिम् अवलोक्य वीणां वादयितुं कथयति। (कवि भगवती भारती (सरस्वती) से यमुना नदी के बेंत की लताओं से युक्त तट पर मधुर मालती की झुकी हुई पंक्तियों को देखकर वीणा बजाने के लिए कहता है।)

3. ‘क’ स्तम्भे पदानि ‘ख’ स्तम्भे तेषां पर्यायपदानि दत्तानि। तानि चित्वा पदानां समक्षे लिखत –
(‘क’ स्तम्भ के पदों के पर्यायपद ‘ख’ स्तम्भ में दिये गये हैं, उनमें से सही पद चुनकर उनके समक्ष लिखिए-)

‘क’ स्तम्भः	‘ख’ स्तम्भः
(क) सरस्वती	1. तीरे
(ख) आम्रम्	2. अलीनाम्
(ग) पवनः	3. समीरः
(घ) तटे	4. वाणी
(ङ) भ्रमराणाम्	5. रसालः

उत्तरम् :

‘क’ स्तम्भः	‘ख’ स्तम्भः
(क) सरस्वती	4. वाणी
(ख) आम्रम्	5. रसालः
(ग) पवनः	3. समीरः
(घ) तटे	1. तीरे
(ङ) भ्रमराणाम्	2. अलीनाम्

4. अधोलिखितानि पदानि प्रयुज्य संस्कृतभाषया वाक्यरचनां कुरुत –
(निम्नलिखित पदों का प्रयोग करते हुए संस्कृत-भाषा में वाक्य-रचना कीजिए)
(क) निनादय
(ख) मन्दमन्दम्
(ग) मारुतः
(घ) सलिलम्
(ङ) सुमनः
उत्तरम् :
वाक्यरचना –
(क) निनादय – अये वाणि! नवीनां वीणां निनादय। (हे सरस्वती! नवीन वीणा को बजाओ।)
(ख) मन्दमन्दम् – कलिन्दात्मजायाः सवानीरतीरे समीरः मन्दमन्दं वहति। (यमुना नदी के बेंत की लताओं से युक्त तट पर हवा धीरे-धीरे चलती है।)
(ग) मारुतः – सायंकाले मारुतः मन्द-मन्दं वहति। (सायंकाल हवा धीरे-धीरे चलती है।)
(घ) सलिलम् – अये वाणि! तव नवीनां वीणाम् आकर्ण्य नदीनां कान्तसलिलं सलीलम् उच्छलेत्। (हे सरस्वती ! तुम्हारी नवीन वीणा को सुनकर नदियों का सुन्दर जल खेल-खेल में उछल पड़े।)
(ङ) सुमनः – पुष्पस्य पर्यायं सुमनः अस्ति। (पुष्प का पर्यायवाची सुमन है।)

5. प्रथमश्लोकस्य आशयं हिन्दीभाषया आङ्गलभाषया वा लिखत।
(प्रथम श्लोक का आशय हिन्दी अथवा अंग्रेजी भाषा में लिखिए।)
उत्तरम् :
प्रथम श्लोक का आशय-हे माँ वाणी (सरस्वती)! आप अपनी वीणा से ऐसे सुन्दर नीतियों से युक्त गीत का मधुर गान करो, जिसे सुनकर समस्त चराचर में सौन्दर्य के प्रति चेतना जाग्रत हो जाए। प्राणी नीति के मार्ग पर अग्रसर हों।

हे माता सरस्वती! वसन्त ऋतु आ गई है। आम के वृक्ष बौरा गए हैं। अमराइयाँ पीली कान्ति से युक्त हो गई हैं। उन आम्र-वृक्षों पर कूकती कोयलों के समूह मनमोहक लगते हैं। फिर भी परतन्त्र भारतीयों के मनों में उत्साह नहीं है। हे माँ सरस्वती ! आप ऐसी वीणा बजाइये, जिससे सभी भारतीयों के मनों में उत्साह भर जाये और वे भारत-माता की स्वतन्त्रता का मार्ग प्रशस्त करें।

6. अधोलिखितपदानां विलोमपदानि लिखत – (निम्नलिखित पदों के विलोमपद लिखिए।)
(क) कठोरम् ………..
(ख) कटु …………
(ग) शीघ्रम् ………..
(घ) प्राचीनम् ……….
(ङ) नीरस: ……….
उत्तरम् :
पदानि – विलोमपदानि
(क) कठोरम् – कोमलम्
(ख) कटु – मृदु
(ग) शीघ्रम् – विलम्बम्
(घ) प्राचीनम् – नवीनम्
(ङ) नीरसः – सरस:

JAC Class 9th Sanskrit भारतीवसन्तगीतिः Important Questions and Answers

प्रश्न: 1.
वाणी किं निनादय? (सरस्वती क्या बजाये?)
उत्तरम् :
वाणी नवीनां वीणां निनादय। (सरस्वती नवीनता से युक्त वीणा को बजाये।)

प्रश्न: 2.
वसन्ते के लसन्ति? (वसन्त में क्या सुशोभित होते हैं ?)
उत्तरम् :
वसन्ते मधुरमञ्जरी-पिञ्जरी-भूतमालाः सरसा: रसाला: लसन्ति। (वसन्त में मधुर मञ्जरियों से पीली हुई सरस आम के वृक्षों की पंक्तियाँ सुशोभित होती हैं।)

प्रश्न: 3.
मन्दमन्दं कः वहति? (धीरे-धीरे क्या बहता है?)
उत्तरम् :
सनीर: समीर: मन्दमन्दं वहति। (जल से युक्त पवन धीरे-धीरे बहता है।)

प्रश्न: 4.
सवानीरतीरः कस्याः अस्ति? (बेंत की लताओं से आच्छादित तट किसका है?)
उत्तरम् :
कलिन्दात्मजायाः सवानीरतीरः अस्ति। (यमुना का तट बेंत की लताओं से आच्छादित है।)

प्रश्नः 5.
पादपाः कैः युक्ताः? (वृक्ष किनसे युक्त हैं?)
उत्तरम् :
पादपाः ललितपल्लवैः युक्ताः सन्ति। (वृक्ष मन को आकर्षित करने वाले पत्तों से युक्त हैं।)

प्रश्नः 6.
अलीनां ततिम् किं करोति? (भ्रमरों की पंक्ति क्या करती है?)
उत्तरम् :
अलीनां ततिम् स्वनन्तीम्। (भ्रमरों की पंक्ति ध्वनि करती है।)

प्रश्न: 7.
किं कृत्वा वीणां निनादयितुं कथितः? (क्या करके वीणा को बजाने के लिए कहा गया है?)
उत्तरम् :
पादपे पुष्पपुजे, मञ्जुकुञ्ज, अलीनां मलिनं ततिं दृष्ट्वा वीणां निनादयितुं कथितः। (वृक्षों पर, पुष्पों के समूह पर, सुन्दर कुञ्जों पर भौंरों की काली पंक्ति अथवा भौरों के काले समूह को देखकर वीणा बजाने को कहा गया है।)

प्रश्न: 8.
कीदृशीं वीणां निनादयितुं कथितः? (कैसी वीणा बजाने के लिए कहा गया है?)
उत्तरम् :
नवीनां वीणां निनादयितुं कथितः। (नवीन वीणा को बजाने के लिए कहा गया है।)

प्रश्न: 9.
वीणाम् आकर्ण्य सुमं किं कुर्यात्? (वीणा को सुनकर पुष्प क्या करे?)
उत्तरम् :
वीणाम् आकर्ण्य सुमं चलेत्। (वीणा को सुनकर पुष्प चलायमान हो जाये।)

प्रश्न: 10.
नदीनां जलं कथम् उच्छलेत्? (नदियों का जल कैसे उछल पड़े?)
उत्तरम् :
नदीनां जलं सलीलम् उच्छलेत्। (नदियों का जल खेल-खेल में उछल पड़े।)

प्रश्न: 11.
वाणी कीदृशीं गीतिं गायतु? (वाणी कैसा गीत गाये?)
उत्तरम् :
वाणी ललित-नीति-लीनाम् गीतिम् गायतु। (वाणी सुन्दर नीति से युक्त गीत गाये।

प्रश्न: 12.
केषां कलापाः लसन्ति? (किनके समूह शोभा दे रहे हैं?)
उत्तरम् :
ललित-कोकिला-काकलीनां कलापाः लसन्ति। (सुन्दर कोमल स्वरों के समूह शोभा देते हैं।)

प्रश्न: 13.
यमुना तीरे केषां नतां पंक्तिमालोक्य वाणी वीणां निनादयतु? (यमुना के किनारे किनकी झुकी हुई पंक्ति को देखकर वाणी वीणा वादन करे।)
उत्तरम् :
मधुमाधवीनां पंक्तिमालोक्य वाणी वीणां निनादयतु। (मधुर माधवी लताओं को देखकर वाणी वीणा बजाये।)

प्रश्न: 14.
पुष्प पुजे केषा ततिः स्वनति? (पुष्पसमूह पर किनकी पंक्ति स्वर करती है ?)
उत्तरम् :
पुष्प पुजे अलीनाम् पक्तिः स्वनति। (पुष्पसमूह पर भौरों की पंक्ति स्वर करता है।)

प्रश्न: 15.
‘भारतीवसन्तगीतिः’ कस्य रचना अस्ति? (‘भारतीवसन्तगीतिः’ किसकी रचना है? )
उत्तरम् :
भारतीवसन्तगीति : पं. जानकी वल्लभ शास्त्रिण: रचना अस्ति। (भारतीवसन्तगीति पं. जानकी वल्लभ शास्त्री की रचना है।

रेखांकित पदान्यधिकृत्य प्रश्न निर्माणं कुरुत। (रेखांकित शब्दों के आधार पर प्रश्न निर्माण कीजिए।)

प्रश्न: 1.
वसन्ते लसन्ति सरसाः रसालाः। (वसन्त में सरस आम शोभा देते हैं।)
उत्तरम् :
वसन्ते के लसन्ति? (वसन्त में कौन शोभा देते हैं?)

प्रश्न: 2.
वाणी नवीनां वीणां निनादयतु? (वाणी नई वीणा बजाये।)
उत्तरम् :
वाणी कीदृशी वीणां निनादयतु? (वाणी कैसी वीणा बजाये?)

प्रश्न: 3.
वहति मन्द-मन्द सनीरः समीरः। (धीरे-धीरे सजल वायु चलती है।)
उत्तरम् :
मन्दम् मन्दम् कीदृशः समीर: वहति? (धीरे-धीरे कैसी हवा चलती है?)

प्रश्न: 4.
उच्छलेत् कान्तं सलिलम्। (सुन्दर जल उछले।)
उत्तरम् :
कीदृशं सलिलं उच्छलेत्। (कैसा पानी उछले?)

प्रश्नः 5.
तवाकर्ण्य वीणामदीनां नदीनां जलं उच्छलेत्। (तुम्हारी ओजस्वी वीणा को सुनकर नदियों का जल उछल पड़े।)
उत्तरम् :
नदीनां जलं काम् आकर्ण्य उच्छलेत्? (नदियों का जल क्या सुनकर उछल पड़े?)

परियोजना-कार्यम्

पाठेऽस्मिन् वीणायाः चर्चा अस्ति। अन्येषां पञ्चवाद्ययन्त्राणां चित्रं रचयित्वा संकलय्य वा तेषां नामानि लिखत।
(इस पाठ में वीणा की चर्चा है। दूसरे पाँच वाद्य-यन्त्रों के चित्र बनाकर अथवा संकलित करके उनके नाम लिखिए।) निर्देश-यहाँ कुछ वाद्य-यन्त्रों के नाम संस्कृत में दिये जा रहे हैं, छात्र इनके चित्र स्वयं बनाएँ या उनका संकलन करें।
वाद्य यन्त्र – (1) वंशी (बाँसुरी) (2) डमरूः (डमरू) (3) ढक्का (ढोल) (4) झिल्लिका (झाँझ) (5) मृदङ्गः (डफली) (6) सारङ्गी (सारंगी) (7) शङ्खः (शंख) (8) घण्टिका (घंटी) (9) जलतरङ्गम् (जलतरंग) (10) करतालः (करताल)।

भारतीवसन्तगीतिः Summary and Translation in Hindi

पाठ-परिचय – ‘भारतीवसन्तगीतिः’ प्रख्यात कवि पं. जानकी वल्लभ शास्त्री की रचना ‘काकली’ नामक गीत-संग्रह से लिया गया है। इसमें कवि ने अपने देश और मातृ-भूमि की स्वतन्त्रता तथा कल्याण की कामना वाणी (सरस्वती) से की है। प्रार्थना करता हुआ कवि कहता है कि हे सरस्वती! ऐसी नवीन वीणा बजाओ जिससे वसन्त ऋतु में मधुर मञ्जरियों से पीली पंक्ति वाले आम के वृक्ष, कोयल का कूजन, वायु का धीरे-धीरे बहना, अमराइयों में काले भौरों का गुंजन और यमुना आदि नदियों का जल अत्यन्त मनमोहक हो जाये तथा तुम्हारी ओजस्विनी वीणा को सुनकर लताओं के नितान्त शान्त सुमन हिलने लगें अर्थात् ऐसी नवीन ओजस्विनी वीणा बजाओ जिससे सृष्टि में नवीन चेतना का संचार हो। इस गीत में कवि का देशानुराग देखने को मिलता है। स्वतन्त्रता-संग्राम की पृष्ठभूमि में लिखा गया यह गीत एक ऐसे वीणा स्वर की कल्पना करता है, जो नवीन चेतना का आह्वान करने के साथ स्वतन्त्रता प्राप्ति के लिए जनसामान्य को प्रेरित करे। अत: यह गीत ओज और माधुर्य गुणों से परिपूर्ण एक पवित्र राग है।

मूलपाठः,अन्वयः,शब्दार्थाः, हिन्दी-अनुवादः, संस्कृत व्यारव्याःअवबोधनकार्यमच

निनादय नवीनामये वाणि! वीणाम्
मृदुं गाय गीति ललित-नीति-लीनाम्।
मधुर-मञ्जरी-पिञ्जरी-भूत-मालाः
वसन्ते लसन्तीह सरसा रसाला:
कलापाः ललित-कोकिला-काकलीनाम्॥1॥ निनादय………….।।

अन्वयः – अये वाणि! नवीनां वीणां निनादय। ललितनीतिलीनां गीतिं मृदुं गाय। इह वसन्ते मधुर-मञ्जरी पिञ्जरी–भूत-मालाः सरसा: लसन्ती। ललित कोकिला काकलीनां कलापाः। नवीनां वीणां निनादय।

शब्दार्था: – निनादय = नितरां वादय (बजाओ), नवीनाम् = नूतनां (नवीन), अये = भो (अरे), वाणि = सरस्वति (हे सरस्वती), वीणाम् = वल्ली (वीणा को), मृदुम् = मधुरं (कोमल), गाय = स्तुहि (गाओ), गीतिम् = गानं (गीत), ललितनीतिलीनाम् सुन्दरनीतिसंलग्नाम् (सुन्दर नीति में लीन), मधुर = चारु (सुन्दर), मञ्जरी = आम्रकुसुम (आम के पुष्प/बौर), पिञ्जरी-भूत-माला: = पीतपङ्क्तयः (पीले वर्ण से युक्त पंक्तियाँ), वसन्ते = वसन्तकाले (वसन्त ऋतु में), . लसन्ति = शोभन्ते (सुशोभित हो रही हैं), इह = अत्र (यहाँ),सरसा = रसपूर्णाः (रस से पूर्ण), रसालाः = आम्राः = (आम के पेड़) कलापाः = समूहाः (समूह), ललित-कोकिला = मनोहरः पिकः (मनमोहक कोयल), काकली = कोकिलानां ध्वनिः (कोयल की आवाज) अये वाणि! = भो माता सरस्वति! (हे माँ सरस्वती), नवीनाम् = नूतनां (नवीन), वीणाम् = वल्ली (वीणा को), निनादय = नितरां वादय (बजाओ)।

हिन्दी अनुवादः

सन्दर्भ – प्रस्तुत गीतांश में कवि ने माँ सरस्वती से वसन्त ऋतु में प्रकृति में नवीन चेतना का संचार करने वाली ओजस्विनी वीणा को बजाने की प्रार्थना की है।

प्रसंग – हे सरस्वती! नवीन वीणा को बजाओ। सुन्दर नीतियों से पूर्ण गीत का मधुर गान करो। इस वसन्त ऋतु में मधुर आम्रपुष्प (बौरों) के कारण पीले वर्ण से युक्त सरस आम के वृक्षों की पंक्तियाँ सुशोभित हो रही हैं। मनमोहक कोयल
की कूक तथा कोयलों के समूह सुन्दर लग रहे हैं। हे सरस्वती! नवीन वीणा को बजाओ।

हिन्दी-अनुवाद – हे सरस्वती! नवीन वीणा को बजाओ। सुन्दर नीतियों से पूर्ण गीत का मधुर गान करो। इस वसन्त
(बौरों) के कारण पीले वर्ण से युक्त सरस आम के वृक्षों की पंक्तियाँ सुशोभित हो रही हैं। मनमोहक कोयल की कूक तथा कोयलों के समूह सुन्दर लग रहे हैं। हे सरस्वती! नवीन वीणा को बजाओ।

संस्कृत व्यारव्याः

सन्दर्भ: – प्रस्तुतो गीतांशोऽस्माकं ‘शेमुषी’ इति पाठ्यपुस्तकस्य ‘भारतीवसन्तगीतिः’ इति पाठात् उद्धनः। य आधुनिक संस्कृत-साहित्ये प्रख्यातस्य कवेः पं. जानकीवल्लभशास्त्रिणः ‘काकली’ इति गीत संग्रहात् सङ्कलितः। (प्रस्तुत गीतांश हमारी शेमुषी पाठ्यपुस्तक के भारतीवसन्तगीति:’ पाठ से लिया गया है, जो आधुनिक संस्कृत-साहित्य में प्रसिद्ध कवि पं. जानकी वल्लभ शास्त्री के ‘काकली’ गीत-संग्रह से संकलित है।)

प्रसंग: – प्रस्तुत गीतांशे कविना माता सरस्वती वसन्तकाले प्रकृतौ नवचेतना सञ्चारिकाम् ओजस्विनी वीणां वादितुं निवेदिता। (प्रस्तुत गीतांश में कवि द्वारा सरस्वती माँ वसन्त काल में प्रकृति में नई चेतना का सञ्चार करने वाली ओजस्वमयी वीणा को बजाने के लिए निवेदन किया जा रहा है।)

व्याख्या: – हे सरस्वति! त्वं नूतनां वल्लकी वादय। रम्यनयस्य मधुरं गीतं गाय। अस्मिन् वसन्त काले मृदुपुष्पैः पीतवर्णानाम् आम्रवृक्षाणां पङ्क्तयः शोभन्ते। मनमोहकानां पिकानां तेषां केकानां च समूहः शोभन्ते! हे सरस्वति! नूतनां . वल्लकी वादय। (हे सरस्वती! तुम नई वीणा को बजाओ! सुन्दर नीति का मधुर गीत गाओ। इस वसन्त काल में कोमल बौर (मंजरी) से युक्त आम के वृक्ष शोभा दे रहे हैं। मनोहर कोयल और उसकी कूक का समूह शोभा दे रहे हैं। हे सरस्वती! नई वीणा बजाओ।)

अवबोधन कार्यम्

प्रश्न 1.
एकपदेन उत्तरत- (एक शब्द में उत्तर दीजिए-)
(क) सरस्वती कीदृशी वीणां निनादयतु? (सरस्वती कैसी वीणा बजाये?)
(ख) वसन्ते कासां कलापाः भवन्ति ? (वसन्त में किनका कलरव होता है?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत – (पूरे वाक्य में उत्तर दीजिए-)
(क) कीदृशीं गीतिं गायतु? (कैसा गीत गाये?)
(ख) वसन्ते काः सरसा: लसन्ति ? (वसन्त में क्या सरस शोभा देती है?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) ‘वादय’ इति पदस्य पर्यायपदं गीतांशात् चुनत। (‘वादय’ पद का पर्याय पद गीतांश से छाँटकर लिखो।)
(ख) ‘नीरस’ पदस्य विलोमपदं गीतांशात् चित्वा लिखत। (‘नीरस’ पद का विलोम-पद गीतांश में से चुनकर
लिखिए।)
उत्तराणि :
(1) (क) नवीनाम् (नई)।
(ख) कोकिला काकलीनाम् (कोयलों की केका ध्वनि)।

(2) (क) ललितनीतिलीनां मृदुगीतिं गायतु। (सुन्दर नीति से युक्त मृदु गीत गायें।)
(ख) वसन्ते मधुर-मञ्जरी-पिञ्जरी-भूत-मालाः सरसाः लसन्ति। (वसन्त में मधुर पीले रंग की मंजरी सरस शोभा देती हैं।)

(3) (क) निनादय ।
(ख) सरस।

वहति मन्दमन्दं सनीरे समीरे
कलिन्दात्मजायास्सवानीरतीरे,
नतां पङ्किमालोक्य मधुमाधवीनाम् ॥2॥ निनादय………….।।

अन्वयः – कलिन्दात्मजायाः सवानीरतीरे सनीरे समीरे मन्दमन्दं वहति। मधुमाधवीनाम् नतां पंक्तिम् आलोक्य अये वाणि! नवीनां वीणां निनादय।

शब्दार्थाः – वहति = चलति (चलती है, बहती है), मन्दमन्दम् = शनैः-शनैः (धीरे-धीरे), सनीरे = सजले (जल से पूर्ण), समीरे = पवने (हवा में), कलिन्दात्मजायाः = यमुनायाः (यमुना नदी के), सवानीरतीरे = वेतसयुक्त तटे (बेंत की लता से युक्त तट पर), नताम् = नतिप्राप्ताम् (झुकी हुई), पक्तिम् = श्रेणिम् (पक्ति को), अवलोक्य = वीक्ष्य (देखकर), मधुमाधवीनाम् = मधुमाधवीलतानाम् (मधुर मालती लताओं को), अये वाणि = ओ माँ सरस्वति! (हे माँ सरस्वती), नवीनाम् = नूतनां (नवीन), वीणाम् = वल्लकी (वीणा को), निनादय = नितरां वादय (बजाओ)।

हिन्दी अनुवादः

सन्दर्भ – प्रस्तुत गीतांश हमारी पाठ्य-पुस्तक ‘शेमुषी’ के ‘भारतीवसन्तगीतिः’ नामक पाठ से उद्धृत है। यह पाठ आधुनिक संस्कृत-साहित्य के प्रख्यात कवि पं. जानकी वल्लभ शास्त्री की रचना ‘काकली’ नामक गीतसंग्रह से संकलित है।

प्रसंग – प्रस्तुत गीतांश में कवि ने सरस्वती से यमुना के तट पर झुकी हुई मधुर मालती की लताओं को देखकर नवीन वीणा को बजाने की प्रार्थना की है। – हिन्दी-अनुवाद-यमुना नदी के बेंत की लताओं से युक्त तट पर जल से पूर्ण वायु धीरे-धीरे बहती है। (उस हवा से) मधुर मालती की लता-पंक्ति को झुकी हुई देखकर हे सरस्वती! नवीन वीणा को बजाओ।

संस्कृत व्याख्याः

सन्दर्भ: – प्रस्तुतो गीतांशोऽस्माकं ‘शेमुषी’ इति पाठ्यपुस्तकस्य ‘भारतीवसन्तगीतिः’ इति पाठात् उद्धृतः। य आधुनिक संस्कृत-साहित्ये प्रख्यातस्य कवेः जानकीवल्लभशास्त्रिण: ‘काकली’ इति गीतसंग्रहात् संकलितः। (प्रस्तुत गीतांश हमारी ‘शेमुषी’ पाठ्यपुस्तक के ‘भारतीवसन्तगीतिः’ पाठ से लिया गया है। यह आधुनिक संस्कृत-साहित्य में प्रसिद्ध कवि पं. जानकी वल्लभ शास्त्री के ‘काकली’ गीत संग्रह से संकलित है।)

प्रसङ्गः – प्रस्तुतगीतांशे कविः यमुनातटे अवनता: मृदु मालती लताः अवलोकस्य नूतनां वल्लवी वादनाय प्रार्थयति। (प्रस्तुत गीतांश में कवि यमुना के तट पर झुकी कोमल मालती लताओं का अवलोकन कर नई वीणा बजाने के लिए प्रार्थना करता है।)

व्याख्या: – नेत्र लताभिः युक्ते यमुना तटे जलसंयुता पवनः शनैः-शनै: वहति । तेन पवनेन मदमालतीलतानां पङ्क्तयोऽवनंता: सन्ति। तान् अवलोक्य हे सरस्वति! त्वं नूतना वीणाम् वल्ली वा वादय। (बेंत की लताओं से युक्त यमुना के किनारे पर जल से युक्त वायु मन्द-मन्द चल रही है। उस पवन से कोमल मालती लताओं की पंक्तियाँ झुकी हुई हैं। उन्हें देखकर हे सरस्वती! तुम नई वीणा को बजाओ।)

अवबोधन कार्यम

प्रश्न 1.
एकपदेन उत्तरत – (एक शब्द में उत्तर दीजिए-)
(क) समीरः कथं वहति? (हवा कैसे चलती है?)
(ख) यमुना कस्याः आत्मजा? (यमुना किसकी बेटी है?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत- (पूरे वाक्य में उत्तर दीजिए-)
(क) यमुना तीरे वायुः कथं वहति? (यमुना-किनारे वायु कैसी चलती है?)
(ख) सरस्वती किमालोक्य वीणां निनादयतु? (सरस्वती क्या देखकर वीणा बजाये?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) ‘सनीरे समीरे’ इति पदयोः किं विशेषण पदम्? (‘सनीरे-समीरे’ में कौनसा विशेषण है?)
(ख) गीतांशे ‘वहति’ क्रियापदस्य कर्तृपदं लिखत। (गीतांश में ‘वहति’ क्रियापद का कर्ता लिखिए।)
उत्तराणि :
(1) (क) मन्दमन्दम्। (धीरे-धीरे)।
(ख) कलिन्दस्य (सूर्य की)।

(2) (क) यमुना तीरे सनीरे समीरे मन्द-मन्दं वहति। (यमुना के किनारे पानी से युक्त वायु धीरे-धीरे चलती है।)
(ख) मधुरमाधवीनां पंक्तिमालोक्य वीणां निनादयतु सरस्वती। (मधुरमाधवी की पंक्तियों को देखकर सरस्वती वीणा बजाये।)

(3) (क) ‘सनीरे’, इति विशेषणपदम्।
(ख) ‘वायुः’ वहति क्रियाया कर्ता।

ललित-पल्लवे पादपे पुष्पपुजे
मलयमारुतोच्चुम्बिते मञ्जुकुञ्ज,
स्वनन्तीन्ततिम्प्रेक्ष्य मलिनामलीनाम् ॥3॥ निनादय…………।।

अन्वयः – ललितपल्लवे पादपे पुष्पपुजे मञ्जुकुञ्जे मलय-मारुतोच्चुम्बिते स्वनन्तीम् अलीनां मलिनां ततिं प्रेक्ष्य अये वाणि! नवीनां वीणां निनादय।

शब्दार्था: – ललित = मनोहर (सुन्दर, मन को आकर्षित करने वाले), पल्लवे = पत्राणि (पत्तों वाले), पादपे = तरौ (वृक्षों पर), पुष्पपुजे = पुष्पसमूहे (पुष्पों के समूह पर), मलयमारुतोच्चुम्बिते = मलयानिलसंस्पृष्टे (चन्दन वृक्ष की सुगन्धित वायु से स्पर्श किये गये), मञ्जुकुञ्ज = शोभनलताविताने (सुन्दर लताओं से आच्छादित स्थान/सुन्दर कुञ्जों पर), स्वनन्तीम् = ध्वनि कुर्वन्तीम् (ध्वनि करती हुई), ततिम् = पंक्तिम् (समूह को), प्रेक्ष्य = दृष्ट्वा (देखकर), मलिनाम् = कृष्णवर्णाम् (मलिन), अलीनाम् = भ्रमराणाम् (भ्रमरों की)।

हिन्दी अनुवादः

सन्दर्भ – प्रस्तुत गीतांश हमारी पाठ्यपुस्तक ‘शेमुषी’ के ‘भारतीवसन्तगीतिः’ नामक पाठ से उद्धृत है। यह पाठ आधुनिक संस्कृत-साहित्य के प्रख्यात कवि पं. जानकीवल्लभ शास्त्री की रचना ‘काकली’ नामक गीतसंग्रह से संकलित है।

प्रसंग – प्रस्तुत गीतांश में कवि ने सरस्वती से मनमोहक कुञ्जों में काले भौंरों की पंक्ति को देखकर नवीन वीणा को . . बजाने की प्रार्थना की है।

हिन्दी-अनुवाद – चन्दन-वृक्ष की सुगन्धित वायु से स्पर्श किये गए, मन को आकर्षित करने वाले पत्तों से युक्त वृक्षों पर, पुष्पों के समूह पर तथा सुन्दर कुञ्जों पर भौंरों की ध्वनि करती हुई पंक्ति समूह को देखकर हे सरस्वती! नवीन वीणा को बजाओ। संस्कृत व्यारव्याः

सन्दर्भ: – प्रस्तुतोऽयं गीतांशोऽस्माकं ‘शेमुषी’ इति पाठ्य-पुस्तकस्य ‘भारतीवसन्तगीतिः’ इति पाठात् उद्धृतः।’ पाठोऽयमाधुनिक-संस्कृत साहित्यस्य प्रख्यातकवेः पं. जानकी वल्लभ शास्त्रिण: ‘काकली’ इति गीति संग्रहात् सङ्कलितः। (प्रस्तुत गीतांश हमारी शेमुषी पाठ्य-पुस्तक के ‘भारतीवसन्तगीतिः’ पाठ से उद्धृत है। यह पाठ आधुनिक संस्कृत-साहित्य के प्रसिद्ध कवि पं. जानकी वल्लभ शास्त्री के काकली गीत-संग्रह से संकलित है।) .

प्रसङ्गः – गीतांशेऽस्मिन् कविः देवी सरस्वती मनमोहकेषु कुञ्जेषु भ्रमराणां श्यामपंक्तिमवलोक्य नूतनां वल्लकी वादयितुं निवेदयति। (इस गीतांश में कवि देवी सरस्वती को मनमोहक कुंजों में भ्रमरों की श्याम पंक्ति को देखकर नई वीणा बजाने के लिए निवेदन करती है।)

व्याख्या: – चन्दन-वृक्षाणां सुरभित-पवनेन स्पृष्टै: मनमोहकैः पत्रैः संयुतेषु वृक्षेषु, पुष्पस्तवकेषु, रम्य कुंजेषु गुञ्जनरतानां मधुकराणां पंक्ति-सम्मर्द चावलोक्य देवि सरस्वति! नूतनां वल्लकी वादय। (चन्दन के वृक्षों की सुगन्धित पवन से छुई हुई, मनमोहक पत्तों से युक्त वृक्षों पर फूलों के गुच्छों पर सुन्दर कुंजों पर गुंजन करने में रत भौंरों की पंक्तियों के समूह को देख हे देवी सरस्वती! तुम नई वीणा को बजाओ।)

अवबोधन कार्यम्

प्रश्न 1.
एकपदेन उत्तरत – (एक शब्द में उत्तर दीजिए-)
(क) अलीनां पंक्तिः केन स्पृष्टा वहति? (भौरों की पंक्ति किससे स्पर्श कर बहती हैं ?)
(ख) केषां पंक्तिमवलोक्य सरस्वती वीणां निनादयतु? (किनकी पंक्ति को देखकर सरस्वती वीणा बजाये?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत- (पूरे वाक्य में उत्तर दीजिए-)
(क) अलीनां पंक्तिः किं कुर्वन्ती अवलोक्य सरस्वती वीणां निनादयतु ? (क्या करती हुई भौरों की पंक्ति को देखकर सरस्वती वीणा बजाये?)
(ख) मलिनामलीनां पंक्ति कुत्र स्वनति? (मलिन भौंरों की पंक्ति कहाँ स्वर करती है?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)
(क) ‘ललितपल्लवे पादपे’ इत्यत्र ‘पादपे’ इति पदस्य विशेषण पदं लिखत। (‘ललितपल्लव पादपे’ में ‘पादपे’ पद का विशेषण लिखिए।)
(ख) ‘निर्मलाम्’ इति पदस्य विलोमार्थक पदं गीतांशात् चित्वा लिखत : (‘निर्मलाम्’ पद का विलोमपद गीतांश से चुन कर लिखिए।)
उत्तराणि :
(1) (क) मलयमरुतेन। (मलयानिल से)।
(ख) मलिनामलीनाम्। (मलिन भौरों की)।

(2) (क) मलिनामलीनां पंक्तिं स्वनन्तीम् अवलोक्य बीणां निनादयतु। (गुंजन करती मलिन भौरों की पंक्ति को
देखकर सरस्वती वीणा बजाये।)
(ख) मलिनामलीनां पंक्ति मञ्जुकुंजे स्वनति। (मलिन भौंरों की पंक्ति सुन्दर कुञ्ज पर स्वर करती है।)

(3) (क) ‘ललितपल्लवे’ इति पदं पादपे पदस्य विशेषणपदम्। (‘ललितपल्लवे’ पद ‘पादपे’ का विशेषण है।)
(ख) मलिनाम्। (कृष्ण वर्ण/मैली)।

लतानां नितान्तं सुमं शान्तिशीलम्
चलेदुच्छलेत्कान्तसलिलं सलीलम्,
तवाकर्ण्य वीणामदीनां नदीनाम् ॥4॥ निनादय….॥

अन्वयः – तव अदीनां वीणाम् आकर्ण्य लतानां नितान्तं शान्तिशीलम् शुमं चलेत्। नदीनां कान्तसलिलं सलीलम् उच्छलेत्। अये वाणि! नवीनां वीणां निनादय।

शब्दार्थाः – लतानाम् = वल्लरीनां (लताओं के/बेलों के), नितान्तम् = अमितं (अत्यधिक), सुमम् = कुसुमम् (पुष्प को), शान्तिशीलम् = शान्तियुक्तम् (शान्ति से युक्त), चलेत् = गच्छेत् (चलायमान हो जाएँ), उच्छलेत् = ऊर्ध्वं गच्छेत् (उच्छलित हो उठे/ उछल पड़े), कान्तसलिलम् = मनोहरजलम् (सुन्दर जल), सलीलम् = क्रीडासहित साथ), तव = ते (तुम्हारी), आकर्ण्य = निशम्य/श्रुत्वा (सुनकर), वीणाम् = वल्ली (वीणा को), अदीनाम् = तेजस्विनी (ओजस्वी), नदीनाम् = सरिताम् (नदियों का)।

हिन्दी अनुवादः

सन्दर्भ – प्रस्तुत गीतांश हमारी पाठ्यपुस्तक ‘शेमुषी’ के ‘भारतीवसन्तगीतिः’ नामक पाठ से उद्धत है। यह पाठ आधुनिक संस्कृत-साहित्य के प्रख्यात कवि पं. जानकीवल्लभ शास्त्री की रचना ‘काकली’ नामक गीतसंग्रह से संकलित है।

प्रसंग – प्रस्तुत गीतांश में कवि ने सरस्वती से प्रकृति में नवीन प्राण फूंक देने वाली ओजस्विनी वीणा को बजाने की प्रार्थना की है।

हिन्दी-अनुवाद – तुम्हारी ओजस्विनी वीणा को सुनकर लताओं के अत्यधिक शान्ति से युक्त पुष्प चलायमान हो जाएँ। नदियों का सुन्दर जल खेल-खेल में उछल पड़े। हे वाणी (सरस्वती)! (ऐसी ओजस्विनी) नवीन वीणा को बजाओ।

संस्कृत व्यारव्याः

सन्दर्भ: – गीतांशोऽयमस्माकं ‘शेमुषी’ इति पाठ्यपुस्तकस्य ‘भारतीवसन्तगीतिः’ इति पाठात् उद्धृत। पाठोऽयमाधुनिक संस्कृत-साहित्यस्य प्रख्यातकवेः पं. जानकी वल्लभ शास्त्रिणः ‘काकली’ इति गीतसंग्रहात् सङ्कलितः। (यह गीतांश हमारी ठ्य-पुस्तक के ‘भारतीवसन्तगीतिः’ पाठ से लिया गया है। यह पाठ आधुनिक संस्कृत साहित्य के प्रसिद्ध कवि पं. जानकी वल्लभ शास्त्री के काकली गीत-संग्रह से सङ्कलित है।)

प्रसङ्गः – प्रस्तुत गीतांशे कविः प्रकृतौ नव प्राण सम्प्रेषणाय देवी सरस्वती प्रति प्रार्थयति। (प्रस्तुत गीतांश में कवि प्रकृति में नये प्राण भरने के लिए देवी सरस्वती से प्राथना करता

व्याख्या:-हे सरस्वति! तव ओजस्विी वल्लकी श्रुत्वा लतानामपि शान्त पुष्पाणि चलायमानानि भवन्तु (गतिशीलाः भवन्तु) नदीनां स्वच्छं जलं क्रीडायामेव उच्चलतु। हे देवि सरस्वति! एवमोजस्विनी वल्ली वादय। (हे सरस्वती ! तुम्हारी
ओजपूर्ण वीणा को सुनकर लताओं के भी शान्त पुष्प चलायमान (गतिशील) हो जायें। नदियों का स्वच्छ जल खेल ही खेल में उछल पड़े। हे देवी सरस्वती ! ऐसी ओजपूर्ण वीणा बजाओ।)

अवबोधन कार्यम्

प्रश्न 1.
एकपदेन उत्तरत- (एक शब्द में उत्तर दीजिए-)
(क) सरस्वत्याः वीणा कीदृशी अस्ति? (सरस्वती की वीणा कैसी है? )
(ख) कासां जलम् उच्छलति? (किनका जल उछलता है?)

प्रश्न 2.
पूर्णवाक्येन उत्तरत- (पूरे वाक्य में उत्तर दीजिए-)
(क) किमाकर्ण्य लता चलति? (क्या सुनकर लता चलायमान होती है?)
(ख) नदीनां स्वच्छं जलं कथम् उच्छलति? (नदियों का स्वच्छ जल कैसे उछलता है?)

प्रश्न 3.
यथानिर्देशम् उत्तरत-(निर्देशानुसार उत्तर दीजिए-)।
(क) ‘नितान्तं शान्तिशीलं सुमम्’ विशेष्य पदं किम्? (विशेष्य पद क्या है?)
(ख) ‘उच्छलेत्’ इति क्रियापदस्य कर्तृपदं गीतांशात् लिखत। (‘उच्छलेत्’ क्रियापद का कर्ता गीतांश से लिखिए।)
उत्तराणि :
(1) (क) अदीना। (ओजपूर्ण) ।
(ख) नदीनाम् (नदियों का)।

(2) (क) सरस्वत्याः अदीनां वीणाम् आकर्ण्य लताः चलन्ति। (सरस्वती की ओजपूर्ण वीणा को सुनकर लताएँ
चलायमान हो जाती हैं।)
(ख) सरस्वत्याः अदीनां वीणां श्रुत्वा नदीनां जलं सलीलम् उच्छलति। (सरस्वती की ओजस्वी वीणा को
सुनकर नदियों का जल उछलता है।)

(3) (क) सुमम् (पुष्प) ।
(ख) ‘जलम्’ उच्छलेत् क्रियापदस्य कर्तृपदम्।

Jharkhand Board JAC Class 9 Maths Important Questions Chapter 7 Triangles Important Questions and Answers.

JAC Board Class 9th Maths Important Questions Chapter 7 Triangles

Question 1.
If D is the mid-point of the hypotenuse AC of a right triangle ABC, prove that BD = \(\frac {1}{2}\)AC.
Solution :
Let ΔABC is a right triangle such that ∠B = 90° and D is midpoint of AC then we have to prove that BD = \(\frac {1}{2}\)AC, we produce BD to E such that BD = DE and join EC

Now, in ΔADB and ΔCDE we have
AD = DC (Given)
BD = DE (By construction)
And, ∠ADB = ∠CDE
[Vertically opposite angles]
∴ By SAS criterion of congruence, we have
ΔADB ≅ ΔCDE
⇒ EC = AB and ∠CED = ∠ABD ….(i)
(By CPCT)
But ∠CED and ∠ABD are alternate interior angles .
∴ CE || AB ⇒ ∠ABC + ∠ECB = 180°
(interior angles)
⇒ 90° + ∠ECB = 180°
⇒ ∠ECB = 90°
Now, in ΔABC and ΔECB, we have
AB = EC [By (i)]
BC = BC [Common]
And, ∠ABC = ∠ECB = 90°
∴ BY SAS criterion of congruence
ΔABC ≅ ΔECB
⇒ AC = EB = 2BD [By CPCT and also D is a midpoint of AC and BE]
⇒ BD = \(\frac {1}{2}\) AC Hence, proved.

Question 2.
In a right-angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side.
Solution :
Let ΔABC is a right triangle such that ∠B = 90° and ∠ACB = 2∠CAB, then we have to prove AC = 2BC. We produce CB to D such that BD = CB and join AD. Let
⇒ ∠ACB = 2x and ∠CAB = x

Proof: In ΔABD and ΔABC, we have
BD = BC (By construction)
AB = AB [Common]
∠ABD = ∠ABC = 90°
∴ By SAS criterion of congruence we get
ΔABD ≅ ΔABC
⇒ AD = AC and ∠DAB = ∠CAB
(By CPCT)
⇒ AD = AC and ∠DAB = x[∵ ∠CAB = x]
Now, ∠DAC = ∠DAB + ∠CAB = x + x = 2x
∴ ∠DAC = ∠ACD
⇒ DC = AD
[Sides opposite to equal angles] (∵ DC = 2BC)
⇒ 2BC = AD
⇒ 2BC = AC (AD = AC)
Hence, Proved.

Question 3.
In figure, T is a point on side QR of ΔPQR and S is a point such that RT = ST. Prove that PQ + PR > QS.

Solution :
In ΔPQR we have
PQ + PR > QR
⇒ PQ + PR > QT + TR
⇒ PQ + PR > QT + ST
[∵ RT = ST]
In ΔQST, QT + ST > SQ
∴ PQ + PR > SQ Hence, proved.

Question 4.
In the given figure, PQ = QR and ∠x = ∠y. Prove that AR = PB.

Solution :
Proof: In the figure, ∠QAR + ∠PAR = 180°(Linear pair axiom)
⇒ ∠QAR + ∠x = 180°
⇒ ∠QAR = 180° – ∠x°
Similarly ∠QBP + ∠RBP = 180° (Linear pair axiom)
⇒ ∠QBP + ∠y = 180°
⇒ ∠QBP = 180° – ∠y …(ii)
But given, ∠x = ∠y
∴ ∠QAR = ∠QBP [From (i) and (ii)]
Now, in ΔQAR and ΔQBP, QR = PQ (Given)
∠QAR = ∠QBP (As proved above)
∠Q = ∠Q (Common)
⇒ ΔQAR = ΔQBP
(AAS congruence rule)
⇒ AR = PB (CPCT)
Hence proved.

Question 5.
Diagonal AC and BD of quadrilateral ABCD intersect each other at O. Prove that
(i) AB + BC + CD + DA > AC + BD
(ii) AB + BC + CD + DA < 2 (AC + BD)

Solution :
Given: AC and BD are the diagonals of quadrilateral ABCD. (i) To prove: AB + BC + CD + DA > AC + BD
Proof: We know that the sum of any two sides of a triangle is always greater than the third side. Therefore,
In ΔABC, AB + BC > AC …(i)
In ΔBCD, BC + CD > BD …(ii)
In ΔCDA CD + DA > CA …(iii)
In ΔABD, AB + AD > BD …(iv)
Adding (i), (ii), (iii) and (iv), we get
2 (AB + BC + CD + DA) > 2 (AC + BD)
⇒ AB + BC + CD + DA > AC + BD
Hence proved.

(i) To prove: AB + BC + CD + DA < 2 (AC + BD) Proof : In ΔOAB, OA + OB > AB ……(i)
In ΔBOC, OB + OC > BC …(ii)
In ΔCOD, OC + OD > CD …(iii)
In ΔAOD, OA + OD > DA …(iv)
Adding (i), (ii), (iii) and (iv), we get
2 (OA + OB + OC + OD) > AB + BC + CD + DA
2 [(OA + OC) + (OB + OD)] > AB + BC + CD + DA
2 (AC + BD) > AB + BC + CD + DA
AB + BC + CD + DA < 2 (AC + BD)
Hence proved

Multiple Choice Questions

Question 1.
If the three altitudes of a Δ are equal then triangle is :
(a) isosceles
(b) equilateral
(c) right-angled
(d) none
Solution :
(b) equilateral

Question 2.
ABCD is a square and P, Q, R are points on AB, BC and CD respectively such that AP = BQ = CR and ∠PQR = 90°, then ∠QPR =
(a) 45°
(b) 50°
(c) 60°
(d) 75°
Solution :
(a) 45°

Question 3.
In a ΔXYZ, LM ⊥ YZ and bisectors YN and ZN of ∠Y and ∠Z respectively meet at Non LM then YL + ZM =
(a) YZ
(b) XY
(c) XZ
(d) LM
Solution :
(d) LM

Question 4.
In a ΔPQR, PS is bisector of ∠P and ∠Q = 70°, ∠R = 30°, then
(a) QS > PQ > PR
(b) QS < PQ < PR
(c) PQ > QS > SR
(d) PQ < QS < SR
Solution :
(b) QS < PQ < PR

Question 5.
If D is any point on the side BC of a ΔABC, then:
(a) AB + BC + CA > 2AD
(b) AB + BC + CA < 2AD
(c) AB + BC + CA > 3AD
(d) None of these
Solution :
(a) AB + BC + CA > 2AD

Question 6.
For given figure, which one is correct:
(a) ΔABC ≅ ΔDEP
(b) ΔABC ≅ ΔFED
(c) ΔABC ≅ ΔDFE
(d) ΔABC ≅ ΔEDF

Solution :
(a) ΔABC ≅ ΔDEP

Question 7.
In a right-angled triangle, one acute angle is double the other then the hypotenuse is :
(a) Equal to the smallest side
(b) Double the smallest side
(c) Triple the smallest side
(d) None of these
Solution :
(d) None of these

Students should go through these JAC Class 9 Maths Notes Chapter 13 Surface Areas and Volumes will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 13 Surface Areas and Volumes

Solid Figures:
If any figure such as cuboids, has three dimensions length, width and height then it is known as three dimensional figures. Rectangle has only two dimensions i.e. length and width. Three dimensional figures have volume in addition to areas of surface from which these solid figures are formed.

→ Cuboid:
There are six faces (rectangular), eight vertices and twelve edges in a cuboid.

Total Surface Area (T.S.A.): The area of surface from which cuboid is formed. If l, b, h be length, breadth and height of a cuboid respectively then
(i) Total Surface Area (T.S.A.) = 2 [l × b + b × h + h × l]
(ii) Lateral Surface Area (L.S.A.) = 2h (b + l)
(or Area of 4 walls) = 2h(l + b)
(iii) Volume of Cuboid = (Area of base) × height = (l × b) × h
(iv) Length of diagonal = \(\sqrt{l^2+b^2+h^2}\)

→ Cube:
Cube has six faces. Each face is a square.

If x is a side of cube, then
(i) T.S.A. = 2[x·x + x·x + x·x]
= 2[x² + x² + x²] = 2(3x²) = 6x²
(ii) L.S.A. = 2[x² + x²] = 4x²
(iii) Length of diagonal = x\(\sqrt{3}\)
(iv) Volume = (Area of base) × Height
= (x²) × x = x³

→ Cylinder:
Curved surface area of cylinder (C.S.A.): It is the area of surface from which the cylinder is formed. When we cut this cylinder, we will find a rectangle with length 2πr and height h units.

If r be the radius and h be height of a cylinder, then
(i) C.S.A. of cylinder = (2πr) × h
= 2πrh.
(ii) T.S.A.C.S.A. + area of circular top and bottom
= 2πrh+ (πr²) + (πr²)
= 2πrh + 2πr² = 2πr(h + r) sq.units
(iii) Volume of cylinder
= Area of base × height
= (πr²) × h = πr²h cubic units

→ Hollow cylinder:

(i) C.S.A. of hollow cylinder
= 2π(R + r)h sq. units
(ii) T.S.A. of hollow cylinder
= 2π(R + r)h + 2π(R² – r²)
= 2π(R + r) [h + R – r] sq. units
(iii) Volume of hollow cylinder
= π(R² – r²)h cubic units
Where, r = inner radius of cylinder
R = outer radius of cylinder
h = height of the cylinder

→ Cone:

(i) C.S.A. of cone = πrl
(ii) T.S.A.of cone = C.S.A. + Base area
πrl + πr² = πr(l + r)
(ii) Volume of cone = \(\frac{1}{3}\)πr²h
Where, h = height of cone
r = radius of base of cone
l = slant height of cone

→ Sphere:

(i) T.S.A. of sphere = 4πr²
(ii) Volume of sphere = \(\frac{4}{3}\)πr3
Where r be the radius of the sphere.

→ Hemisphere:

(i) C.S.A. = 2πr²
(ii) T.S.A. = C.S.A. + base area
= 2πr² + πr² = 3πr²
(iii) Volume = \(\frac{2}{3}\)πr³
Where r is the radius of Hemisphere

→ Hollow Hemisphere:

(i) C.S.A. = 2π(R² + r²)
(ii) T.S.A. = 2π(R² + r²) + π(R² – r²)
(iii) Volume = \(\frac{2}{3}\)π(R³ – r³)

Jharkhand Board JAC Class 9 Maths Important Questions Chapter 6 Lines and Angles Important Questions and Answers.

JAC Board Class 9th Maths Important Questions Chapter 6 Lines and Angles

Question 1.
Two supplementary angles are in the ratio 4 : 5, find the angles.
Solution :
Let angles be 4x and 5x.
∵ Angles are supplementary
∴ 4x + 5x = 180°
⇒ 9x = 180°
⇒ x = \(\frac {180°}{2}\) = 20°
∴ Angles are 4 × 20° and 5 × 20° i.e., 80° and 100°

Question 2.
If an angle differs from its complement by 10°, find the angle.
Solution :
Let angle be x° then its complement is 90° – x°.
Now given, x° – (90° – x°) = 10°
⇒ x° – 90° + x° = 10°
⇒ 2x° = 10° + 90° = 100°
⇒ x° = \(\frac {100°}{2}\) = 50°
∴ Required angle is 50°.

Question 3.
In figure, OP and OQ bisects ∠BOC and ∠AOC respectively. Prove that ∠POQ = 90°.

Solution :
∵ OP bisects ∠BOC
∴ ∠POC = \(\frac {1}{2}\)∠BOC ………(i)
Also OQ bisects ∠AOC
∠COQ = \(\frac {1}{2}\)∠AOC ………(ii)
∵ OC stands on AB
∴ ∠AOC + ∠BOC = 180°[Linear pair]
⇒ \(\frac {1}{2}\)∠AOC + \(\frac {1}{2}\)∠BOC = \(\frac {1}{2}\) × 180°
⇒ ∠COQ + ∠POC = 90° [Using (i) and (ii)]
⇒ ∠POQ = 90°

Question 4.
In figure, lines AB, CD and EF intersect at O. Find the measures of ∠AOC, ∠DOE and ∠BOF.

Solution :
Given, ∠AOE = 40° and ∠BOD = 35°
Clearly, ∠AOC = ∠BOD
(Vertically opposite angles)
⇒ ∠AOC = 35°
⇒ ∠BOF = ∠AOE
[Vertically opposite angles]
⇒ ∠BOF = 40°
Now, ∠AOB = 180° [Straight angle]
⇒ ∠AOC + ∠COF + ∠BOF = 180°
[Angles sum property]
⇒ 35° + ∠COF + 40° = 180°
⇒ ∠COF = 180° – 75° = 105°
Now, ∠DOE = ∠COF
[Vertically opposite angles]
∴ ∠DOE = 105°

Question 5.
In figure if l || m, n || p and ∠1 = 85° find ∠2.

Solution :
∵ n || p and m is transversal
∴ ∠1 = ∠3 = 85° [Corresponding angles]
Also, m || l and p is transversal
∴ ∠2 + ∠3 = 180°
[Consecutive interior angles]
⇒ ∠2 + 85° = 180°
⇒ ∠2 = 180° – 85°
⇒ ∠2 = 95°

Multiple Choice Questions

Question 1.
If two lines are intersected by a transversal, then each pair of corresponding angles so formed is
(a) Equal
(b) Complementary
(c) Supplementary
(d) None of these
Solution :
(a) Equal

Question 2.
Two parallel lines have:
(a) a common point
(b) two common points
(c) no common point
(d) infinite common points
Solution :
(c) no common point

Question 3.
An angle is 14° more than its complement then angle is:
(a) 38°
(b) 52°
(c) 50°
(d) none of these
Solution :
(a) 38°

Question 4.
The angle between the bisectors of two adjacent supplementary angles is:
(a) acute angle
(b) right angle
(c) obtuse angle
(d) none of these
Solution :
(c) obtuse angle

Question 5.
If one angle of triangle is equal to the sum of the other two then triangle is:
(a) acute triangle
(b) obtuse triangle
(c) right triangle
(d) none of these
Solution :
(c) right triangle

Question 6.
Point x is in the interior of ∠BAC. If ∠BAC = 70° and ∠BAX = 42° then ∠XAC =
(a) 28°
(b) 29°
(c) 27°
(d) 30°
Solution :
(a) 28°

Question 7.
If the supplement of an angle is three times its complement, then angle is:
(a) 40°
(b) 35°
(c) 50°
(d) 45°
Solution :
(d) 45°

Question 8.
Two angles whose measures are a and b are such that 2a – 3b = 60° then \(\frac {4a}{5b}\) = _______, if they form a linear pair :
(a) 0
(b) 8/5
(c) \(\frac {12}{5}\)
(d) \(\frac {2}{3}\)
Solution :
(c) \(\frac {12}{5}\)

Question 9.
Which one of the following statements is not false?
(a) If two angles are forming a linear pair, then each of these angles is of measure 90°
(b) Angles forming a linear pair can both be acute angles.
(c) One of the angles forming a linear pair can be obtuse angle.
(d) Bisectors of the adjacent angles form a right angle.
Solution :
(c) One of the angles forming a linear pair can be obtuse angle.

Question 10.
Which one of the following is correct?
(a) If two parallel lines are intersected by a transversal, then alternate interior angles are equal
(b) If two parallel lines are intersected by a transversal then sum of the interior angles on the same side of transversal is 180°
(c) If two parallel lines are intersected by a transversal then corresponding angles are equal.
(d) All of these
Solution :
(d) All of these

Jharkhand Board JAC Class 9 Maths Important Questions Chapter 9 Areas of Parallelograms and Triangles Important Questions and Answers.

JAC Board Class 9th Maths Important Questions Chapter 9 Areas of Parallelograms and Triangles

Question 1.
In a parallelogram ABCD; AB = 8 cm. The altitudes corresponding to sides AB and AD are respectively 4 cm and 5 cm. Find AD.

Solution :
We know that, Area of a parallelogram
= Base × Corresponding altitude
∴ Area of parallelogram
ABCD = AD × BN = AB × DM
⇒ AD × 5 = 8 × 4
⇒ AD = \(\frac{8 \times 4}{5}\) = 6.4 cm.

Question 2.
ABCD is a quadrilateral and BD is one of its diagonals as shown in the figure. Show that the quadrilateral ABCD is a parallelogram and find its area.

Solution :
From figure, the transversal DB is intersecting a pair of lines DC and AB such that
∠CDB = ∠ABD = 90°.
As these angles form a pair of alternate interior angles
∴ DC || AB.
Also, DC = AB = 2.5 units.
∴ Quadrilateral ABCD is a parallelogram. Now, area of parallelogram ABCD
= Base × Corresponding altitude
= 2.5 × 4 = 10 sq. units

Question 3.
In figure, E is any point on median AD of a ΔABC. Show that ar(ABE) = ar(ACE).

Solution :
Construction: From A, draw AG ⊥ BC and from E, draw EF ⊥ BC.
Proof: ar(ΔABD) = \(\frac {BD × AG}{2}\)
ar(ΔADC) = \(\frac{\mathrm{DC} \times \mathrm{AG}}{2}\)
But, BD = DC [∵ D is the mid-point of BC as AD is the median]
∴ ar(ΔABD) = ar(ΔADC) ……………(i)
Again, ar(ΔEBD) = \(\frac{\mathrm{BD} \times \mathrm{EF}}{2}\)
ar(ΔEDC) = \(\frac{\mathrm{DC} \times \mathrm{EF}}{2}\)
But, BD = DC
∴ ar(ΔEBD) = ar(ΔEDC) …….(ii)
Subtracting (ii) from (i), we get
ar(ΔABD) – ar(ΔEBD)
= ar(ΔADC) – ar(ΔEDC)
⇒ ar(ΔABE) = ar(ΔACE)
Hence, proved.

Question 4.
Triangles ABC and DBC are on the same base BC; with A, D on opposite sides of the line BC, such that ar(ΔABC) = ar(ΔDBC). Show that BC bisects AD.

Solution :
Construction: Draw AL ⊥ BC and DM ⊥ BC.
Proof: ar(ΔABC) = ar(ΔDBC) (Given)
⇒ \(\frac{\mathrm{BC} \times \mathrm{AL}}{2}=\frac{\mathrm{BC} \times \mathrm{DM}}{2}\)
⇒ AL = DM ………….(i)
Now in Δs OAL and OMD
AL = DM [From (i)]
⇒ ∠ALO = ∠DMO [Each = 90°]
⇒ ∠AOL = ∠MOD [Vert. opp. ∠s]
∴ ΔOAL ≅ ΔODM [By AAS]
∴ OA = OD [By CPCT]
i.e., BC bisects AD. Hence, proved.

Question 5.
ABC is a triangle in which D is the mid-point of BC and E is the mid-point of AD. Prove that the area of ΔBED = \(\frac {1}{4}\) area of ΔABC.

Solution :
Given: A ΔABC in which D is the midpoint of BC and E is the mid-point of AD.
To prove: ar(ΔBED) = \(\frac {1}{4}\)ar(ΔABC).
Proof: ∵ AD is a median of ΔABC
∴ ar(ΔABD) = ar(ΔADC)
= \(\frac {1}{2}\)ar(ΔABC) ……..(i)
[∵ Median of a triangle divides it into two triangles of equal area]
Again,
∵ BE is a median of ΔABD.
∴ ar(ΔBEA) = ar(ΔBED)
= \(\frac {1}{2}\)ar(ΔABD) ……..(ii)
[∵ Median of a triangle divides it into two triangles of equal area]
∴ ar(ΔBED) = \(\frac {1}{2}\)ar(ΔABD)
= \(\frac {1}{2}\) × \(\frac {1}{2}\)ar(ΔABC) [From (i)]
∴ ar(ΔBED) = \(\frac {1}{4}\)a(ΔABC). [From (ii)]

Question 6.
Prove that the area of an equilateral triangle is equal to \(\frac{\sqrt{3}}{4}\)a², where a is the side of the triangle.

Solution :
Draw AD ⊥ BC
⇒ ΔABD ≅ ΔACD
[By RHS congruence rule]
∴ BD = DC [By CPCT]
∴ BD = DC = \(\frac {a}{2}\)
In right-angled ΔABD
AD² = AB² – BD²
= a² – (\(\frac {a}{2}\))² = a² – \(\frac{a^2}{4}=\frac{3 a^2}{4}\)
AD = \(\frac{\sqrt{3} a}{2}\)
Area of ΔABC = \(\frac {1}{2}\)BC × AD
= \(\frac {1}{2}\)a × \(\frac{\sqrt{3} a}{2}=\frac{\sqrt{3} a^2}{4}\)
Hence, proved.

Question 7.
In figure, P is a point in the interior of rectangle ABCD. Show that
(i) ar(ΔAPD) + ar(ΔBPC)
= \(\frac {1}{2}\)ar(rect. ABCD)

(ii) ar(APD) + ar(PBC)
= ar(APB) + ar(PCD)

Solution :
Given: A rect. ABCD and P is a point inside it. PA, PB, PC and PD have been joined.
To prove:
(i) ar(ΔAPD) + ar(ΔBPC)
= \(\frac {1}{2}\)ar(rect. ABCD)

(ii) ar(ΔAPD) + ar(ΔBPC)
= ar(ΔAPB) + ar(ΔCPD).
Construction :
Draw EPF || AB
and LPM || AD.
Proof: EPF || AB and DA cuts them,

(i) ∠EAB = 90° [As each angle of a rectangle is of 90°]
∴ ∠DEP = ∠EAB = 90° [Corresponding angles]
∴ PE ⊥ AD
Similarly, PF ⊥ BC; PL ⊥ AB and PM ⊥ DC.
∴ ar(ΔAPD) + ar(ΔBPC)
= (\(\frac {1}{2}\) × AD × PE) + ar (\(\frac {1}{2}\) × BC × PF)
= \(\frac {1}{2}\)AD (PE + PF) [∵ BC = AD]
= \(\frac {1}{2}\) × AD × EF = \(\frac {1}{2}\) × AD × AB [∵ EF = AB]
= \(\frac {1}{2}\) × ar(rectangle ABCD)

(ii) ar(ΔAPB) + ar(PCD)
= (\(\frac {1}{2}\) × AB × PL) + (\(\frac {1}{2}\) × DC × PM)
= \(\frac {1}{2}\) × AB × (PL + PM) [∵ DC = AB]
= \(\frac {1}{2}\) × AB × LM
= \(\frac {1}{2}\) × AB × AD [∵ LM = AD]
= \(\frac {1}{2}\) × ar(rect. ABCD).
= ar(ΔAPD) + ar(PBC)
= ar(ΔAPB) + ar(PCD)
Hence, proved.

Multiple Choice Questions

Question 1.
The sides BA and DC of the parallelogram ABCD are produced as shown in the figure then

(a) a + x = b + y
(b) a + y = b + a
(c) a + b = x + y
(d) a – b = x – y
Solution :
(c) a + b = x + y

Question 2.
The sum of the interior angles of polygon is three times the sum of its exterior angles. Then numbers of sides in polygon is
(a) 6
(b) 7
(c) 8
(d) 9
Solution :
(d) 9

Question 3.
In the following figure, AP and BP are angle bisectors of ∠A and ∠B which meet at a point P of the parallelogram ABCD. Then 2∠APB =

(a) ∠A + ∠B
(b) ∠A + ∠C
(c) ∠B + ∠D
(d) 2∠C + ∠D
Solution :
(a) ∠A + ∠B

Question 4.
In a parallelogram ABCD, AO and BO are respectively the angle bisectors of ∠A and ∠B (see figure). Then measure of ∠AOB is

(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution :
(d) 90°

Question 5.
In a parallelogram ABCD, ∠D = 60° then the measurement of ∠A is
(a) 120°
(b) 65°
(c) 90°
(d) 75°
Solution :
(a) 120°

Question 6.
In the adjoining figure ABCD, the angles x and y are

(a) 60°, 30°
(b) 30°, 60°
(c) 45°, 45°
(d) 90°, 90°
Solution :
(a) 60°, 30°

Question 7.
In the parallelogram PQRS (see figure), the values of ∠SQP and ∠QSP are

(a) 45°, 60°
(b) 60°, 45°
(c) 70°, 35°
(d) 35°, 70°
Solution :
(a) 45°, 60°

Question 8.
In parallelogram ABCD, AB = 12 cm. The altitudes corresponding to the sides CD and AD are respectively 9 cm and 11 cm. Find AD.

(a) \(\frac {108}{11}\) cm
(b) \(\frac {108}{10}\) cm
(c) \(\frac {99}{10}\) cm
(d) \(\frac {108}{17}\) cm
Solution :
(a) \(\frac {108}{11}\) cm

Question 9.
In ΔABC, AD is a median and P is a point on AD such that AP : PD = 1 : 2 then the area of ΔABP =
(a) \(\frac {1}{2}\) × Area of ΔABC
(b) \(\frac {2}{3}\) × Area of ΔABC
(c) \(\frac {1}{3}\) × Area of ΔABC
(d) \(\frac {1}{6}\) × Area of ΔABC
Solution :
(d) \(\frac {1}{6}\) × Area of ΔABC

Question 10.
In ΔABC if D is a point on BC and divides it in the ratio 3 : 5 i.e., if BD : DC = 3 : 5 then, ar(AADC): ar(ΔABC) = ?
(a) 3 : 5
(b) 3 : 8
(c) 5 : 8
(d) 8 : 3
Solution :
(b) 3 : 8

Students should go through these JAC Class 9 Maths Notes Chapter 12 Heron’s Formula will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 12 Heron’s Formula

Mensuration:
A branch of mathematics which concerns itself with measurement of lengths, areas and volumes of plane and solid figure is called Mensuration.
→ Perimeter: The perimeter of a plane figure is the length of its boundary. In case of a triangle or a polygon, the perimeter is the sum of the lengths of its sides.

→ Units of Perimeter: The unit of perimeter is the same as the unit of length i.e. centimetre (cm), metre (m), kilometre (km) etc.
1 centimetre (cm) = 10 milimetres (mm)
1 decimetre (dm) = 10 centimetres
1 metre (m) = 10 decimetres
= 100 centimetres
= 1000 milimetres
1 decametre (dam) = 10 metres
= 1000 centimetres
1 hectometre (hm) = 10 decametres
= 100 metres
1 kilometre (km) = 1000 metres
= 100 decametres
= 10 hectometres

Area:
The area of a plane figure is the measure of the surface enclosed by its boundary.
The area of a triangle or a polygon is the measure of the surface enclosed by its sides.
→ Units of Area:
The various units of measuring area are, square centimetre (cm²), square metre (m²), hectare etc.
1 square centrimetre (cm²)
= 1 cm × 1 cm
= 10 mm × 10 mm
= 100 sq.mm.
1 square decimetre (dm²)
= 1 dm × dm
= 10 cm × 10 cm
= 100 sq. cm.
1 square metre (m²)
= 1 m × 1 m
= 10 dm × 10 dm
= 100 sq. dm
1 square decametre (dam²)
= 1 dam × 1 dam
= 10 m × 10 m
= 100 sq.m.
1 square hectometre (hm²)
= 10 dam × 10 dam
= 100 sq. dam
(or 1 hectare) = 10000 sq. m.
1 square kilometre (km²)
= 1 km × 1 km
= 10 hm × 10 hm
= 100 sq. hm.

→ Heron’s formula:

In ΔABC if sides of triangle BC, CA, and AB are a, b, c respectively then
perimeter = 2s = a + b + c
Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

→ Perimeter and Area of a Triangle:
Right-angled triangle:

For a right-angled triangle, let b be the base, h be the perpendicular and d be the hypotenuse. Then
(A) Perimeter = b + h + d
(B) Area = \(\frac{1}{2}\)(Base × Height) = \(\frac{1}{2}\)bh
(C) Hypotenuse, d = \(\sqrt{b^2+h^2}\) [Pythagoras theorem]

Isosceles right-angled triangle:

For an isosceles right-angled triangle, let a be the equal sides, then
(A) Hypotenuse = \(\sqrt{a^2+a^2}=\sqrt{2} a\)
(B) Perimeter = 2a + \(\sqrt{2}\)a
(C) Area = \(\frac{1}{2}\)(Base × Height) = \(\frac{1}{2}\)(a × a) = \(\frac{1}{2}\)a².

Equilateral triangle:

For an equilateral triangle, let each side be a, and the height of the triangle be h, then
(A) ∠A = ∠B = ∠C = 60°
(B) ∠BAD = ∠CAD = 30°
(C) AB = BC = AC = a(say)
(D) BD = DC = \(\frac{\mathrm{a}}{2}\)
(E) \(\left(\frac{\mathrm{a}}{2}\right)^2\) + h² = a² ⇒ h² = \(\frac{3 \mathrm{a}^2}{4}\)
∴ Height (h) = \(\frac{\sqrt{3}}{2} \mathrm{a}\)
(F) Area = \(\frac{1}{2}\)(Base × Height)
= \(\frac{1}{2} \times \mathrm{a} \times \frac{\sqrt{3}}{2} \mathrm{a}=\frac{\sqrt{3}}{4} \mathrm{a}^2\)
(G) Perimeter = a + a + a = 3a.

Jharkhand Board JAC Class 9 Maths Important Questions Chapter 11 Constructions Important Questions and Answers.

JAC Board Class 9th Maths Important Questions Chapter 11 Constructions

Question 1.
Construct an equilateral triangle if its altitude is 3.2 cm.
Solution :
Given: In an equilateral ΔABC, an altitude
AD = 3.2 cm
Required: To Construct an equilateral triangle ABC from the given data.
STEPS:
(i) Draw a line PQ
(ii) Construct a perpendicular bisector DE to PO.
(iii) Cut off DA = 3.2 cm from DE.
(iv) Construct ∠DAR = 30°.
The ray AR intersects PQ at B.
(v) Similarly, draw ∠DAC = 30.
The ray AC intersects PQ at C.
(vi) Join A with B and C.
We get the required ΔABC.

Question 2.
Construct a right-angled triangle whose hypotenuse measures 8 cm and one side is 6 cm.
Solution :
Given: Hypotenuse AC of a ΔABC = 8 cm and one side AB = 6 cm.
Required: To construct a right-angled ΔABC from the given data.
STEPS:
(i) Draw a line segment AC = 8 cm.
(ii) Mark the mid-point 0 of AC by doing perpendicular bisector of AC.
(iii) With O as centre and radius OA, draw a semicircle on AC.
(iv) With A as centre and radius equal to 6 cm, draw an arc, cutting the semicircle a B.
(v) Join A and B, B and C.
We get the required right-angled triangle ABC

Question 3.
Construct a ΔABC in which BC = 6.4 cm, altitude from A is 3.2 cm and the median bisecting BC is 4 cm.
Solution :
Given: One side BC = 6.4 cm, altitude AD = 3.2 cm and the median AL = 4 cm.
Required: To construct a ΔABC form the given data
STEPS:
(i) Draw BC = 6.4 cm
(ii) Bisect BC at L.
(iii) Draw EF || BC at a distance 3.2 cm for BC
(iv) With L as centre and radius equal to 4 cm, draw an arc, cutting EF at A
(v) Join A and B ; A and C, A and L.
We get the required triangle ABC

Question 4.
Construct a ΔABC in which ∠B = 30° and ∠C = 60° and the perpendicular from the vertex A to the base BC is 4.8 cm.
Solution :
Given: ∠B = 30°, ∠C = 60°, length of perpendicular from vertex A to be base BC = 4.8 cm.
Required: To construct a ΔABC from the given data.
STEPS :
(i) Draw any line PQ.
(ii) Take a point B on line PQ and construct ∠QBR = 30°
(iii) Draw a line EF || PQ at a distance of 4.8 cm from PQ, cutting BR at A.
(iv) Construct an angle ∠FAC = 60°, cutting PQ at C.
(v) Join A and C.
We get the required triangle ABC.

Question 5.
Construct a triangle ABC, the lengths of whose medians are 6 cm, 7 cm and 6 cm.
Solution :
Given: Median AD = 6 cm, median BE = 7 cm, median CF = 6 cm.
Required: To construct a AABC from the given data.
STEPS:
(i) Construct a ΔAPQ with AP = 6 cm, PQ = 7 cm and AQ = 6 cm.
(ii) Draw the medians AE and PF of ΔAPQ intersecting each other at G.
(iii) Produce AE to B such that GE = EB
(iv) Join B and Q and produce it to C, such that BQ = QC
(v) Join A and C. We get the required triangle ABC.