RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2 (Updated for 2021-22)

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• RD Sharma Class 9 Maths Solutions

Download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2

RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2

Access RD Sharma Class 9 Solutions Chapter 13 Exercise 13.2

Question 1.
Two opposite angles of a parallelogram are (3x- 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.
Solution:
∵ Opposite angles of a parallelogram are equal
∴ 3x – 2 = 50 – x
⇒ 3x + x – 50 + 2
⇒ 4x = 52
⇒ x = 524 = 13
∴ ∠A = 3x – 2 = 3 x 13 – 2 = 39° – 2 = 37°
∠C = 50° -x = 50° – 13 = 37°
But∠A + B = 180°
∴ 37° + ∠B = 180°
⇒ ∠B = 180° – 37° = 143°

and ∠D = ∠B (Opposite angles of a ||gm)
∴ ∠D = 143°
Hence angles and 37°, 143°, 37°, 143°

Question 2.
If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.
Solution:
Let in ||gm ABCD,
∠A =x
Then ∠B = 23 x

But, ∠A + ∠B = 180° (Sum of two adjacent angles of a ||gm)
⇒ x + 23x = 180°
⇒ 53x = 180°
⇒ x = 180° x 35 = 108°
∴ ∠A = 108°
and ∠B = 108° x 23 = 72°
But, ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 108°, ∠D = 72°
Hence angles are 108°, 72°, 108°, 72°

Question 3.
Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.
Solution:
Let smallest angle of a ||gm = x
Then second angle = 2x – 24°
But these are consecutive angles
∴ x + (2x- 24°) = 180°
⇒ x + 2x – 24° = 180°

⇒ 3x = 180° + 24° = 204°
⇒ x =204∘3  = 68°
∴ Smallest angle = 68°
and second angle = 2x 68° – 24°
= 136°-24° = 112°
∵ The opposite angles of a ||gm are equal Other two angles will be 68° and 112°
∴ Hence angles are 68°, 112°, 68°, 112°

Question 4.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?
Solution:
In a ||gm ABC,

Perimeter = 22cm
and longest side = 6.5 cm
Let shorter side = x
∴ 2x (6.5 + x) = 22
⇒ 13 + 2x = 22
⇒ 2x = 22 – 13 = 9
⇒ x = 92 = 4.5
∴ shorter side = 4.5cm

Question 5.
In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.
Solution:
In ||gm ABCD,
∠D = 135°
But, ∠A + ∠D = 180° (Sum of consecutive angles)
⇒∠A+ 135° = 180°

⇒ ∠A = 180° – 135° = 45°
∵ ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠B = 135°

Question 6.
ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.
Solution:
In ||gm ABCD,
∠A = 70°
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70° = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles of a ||gm)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 7.
In the figure, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.

Solution:
In ||gm ABCD,
∠A + ∠B = 180°
(Sum of consecutive angles) But, ∠A = 75°

∴ ∠B = 180° – ∠A = 180° – 75° = 105°
∴ DBA = 105° -60° = 45°
But ∠CDB = ∠DBA (alternate angles)
= 45°
and ∠ADB = ∠DBC = 60°

Question 8.
Which of the following statements are true (T) and which are false (F)?
(i) In a parallelogram, the diagonals are equal.
(ii) In a parallelogram, the diagonals bisect each other.
(iii) In a parallelogram, the diagonals intersect each other at right angles.
(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.
(v) If all the angles of a quadrilateral are equal, it is a parallelogram.

(vi) If three sides of a quadrilateral are equal, it is a parallelogram.
(vii) If three angles of a quadrilateral are equal, it is a parallelogram.
(viii)If all the sides of a quadrilateral are equal it is a parallelogram.
Solution:

(i) False, Diagonals of a parallelogram are not equal.
(ii) True.
(iii) False, Diagonals bisect each other at right angles is a rhombus or a square only.
(iv) False, In a quadrilateral, if opposite sides are equal and parallel, then it is a ||gm.
(v) False, If all angles are equal, then it is a square or a rectangle.

(vi) False, If opposite sides are equal and parallel then it is a ||gm
(vii) False, If opposite angles are equal, then it is a parallelogram.
(viii)False, If all the sides are equal then it is a square or a rhombus but not parallelogram.

Question 9.
In the figure, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC= BC and DC = 2AD.

Solution:
Given : In ||gm ABCD,
∠A = 60°
Bisector of ∠A and ∠B meet at P.
To prove :
(i) AD = DP
(ii) PC = BC
(iii) DC = 2AD
Construction : Join PD and PC
Proof : In ||gm ABCD,
∠A = 60°
But ∠A + ∠B = 180° (Sum of excutive angles)
⇒ 60° + ∠B = 180°
∴ ∠B = 1809 – 60° = 120°
∵ DC || AB
∠PAB = ∠DPA (alternate angles)
⇒ ∠PAD = ∠DPA (∵ ∠PAB = ∠PAD)
∴ AB = DP
(PA is its angle bisector, sides opposite to equal angles)

(ii) Similarly, we can prove that ∠PBC = ∠PCB (∵ ∠PAB = ∠BCA alternate angles)
∴ PC = BC
(iii) DC = DP + PC
= AD + BC [From (i) and (ii)]
= AD + AB = 2AB (∵BC = AD opposite sides of the ||gm)
Hence DC = 2AD

Question 10.
In the figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove thatAF = 2AB.
Solution:
Given : In ||gm ABCD,
E a mid point of BC
DE is joined and produced to meet AB produced at F

To prove : AF = 2AB
Proof : In ∆CDE and ∆EBF
∠DEC = ∠BEF (vertically opposite angles)
CE = EB (E is mid point of BC)
∠DCE = ∠EBF (alternate angles)
∴ ∆CDE ≅ ∆EBF (SAS Axiom)
∴ DC = BF (c.p.c.t.)
But AB = DC (opposite sides of a ||gm)
∴ AB = BF
Now, AF = AB + BF = AB + AB = 2AB
Hence AF = 2AB

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