## RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2

Other Exercises

- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
- RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

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 = \(\frac { 52 }{ 4 }\) = 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 = \(\frac { 2 }{ 3 }\) x

But, ∠A + ∠B = 180° (Sum of two adjacent angles of a ||gm)

⇒ x + \(\frac { 2 }{ 3 }\)x = 180°

⇒ \(\frac { 5 }{ 3 }\)x = 180°

⇒ x = 180° x \(\frac { 3 }{ 5 }\) = 108°

∴ ∠A = 108°

and ∠B = 108° x \(\frac { 2 }{ 3 }\) = 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 =\(\frac { { 204 }^{ \circ } }{ 3 }\) = 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 = \(\frac { 9 }{ 2 }\) = 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

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2 are helpful to complete your math homework.

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