# Class 9 RD Sharma Solutions – Chapter 14 Quadrilaterals – Exercise 14.2

**Question 1. Two opposite angles of a parallelogram are (3x – 2)Â° and (50 – x)**Â°**. Find the measure of each angle of the parallelogram.**

**Solution:**

Given: Two opposite angles of a parallelogram are (3x â€“ 2)Â° and (50 â€“ x)Â°.

(3x – 2)Â°= (50 – x)Â° [Opposite sides of a parallelogram are equal]

3x + x = 50 + 2

4x = 52

x = 13

Angle x is 13Â°

(3x – 2) = (3*13 – 2) = 37Â°

(50 – x)Â° = (50 – 13)Â°= 37Â°

x + 37Â°= 180Â° [Adjacent angles of a parallelogram are supplementary]

x = 180Â°âˆ’ 37Â°= 143Â°

Hence, required angles are : 37Â°, 143Â°, 37Â°, 143Â°.

**Question 2. If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.**

**Solution:**

Let the measure of the angle be x.

Therefore, the measure of the angle adjacent is 2x/3

Hence, x + 2x/3 = 180Â° [Consecutive angles of a parallelogram are supplementary]

2x + 3x = 540Â°

5x = 540Â°

x = 108Â°

Now,

âŸ¹ x + 108Â°= 180Â° [Consecutive angles of a parallelogram are supplementary]

âŸ¹ x + 108Â°= 180Â°

âŸ¹ x = 180Â°- 108Â°= 72Â°

âŸ¹ x = 72Â°

required angles are180Â°, 72Â°, 180Â°, 72Â°

**Question 3. Find the measure of all the angles of a parallelogram, if one angle is 24**Â°**less than twice the smallest angle.**

**Solution:**

x + 2x – 24Â°= 180Â° [Consecutive angles of a parallelogram are supplementary]

3x – 24Â°= 180Â°

3x = 108Â° + 24Â°

3x = 204Â°

x = 204/3 = 68Â°

x = 68Â°

Other angle of parallelogram=2x – 24Â°= 2*68Â°- 24Â°= 112Â°

required angles are 68Â°,112Â°,68Â°,112Â°

**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:**

Given: perimeter of a parallelogram is 22 cm

Let us consider the shorter side be ‘y’.

perimeter = y + 6.5 + 6.5 + x [Sum of all sides]

22 = 2(y + 6.5)

11 = y + 6.5

y = 11 – 6.5 = 4.5 cm

Hence, the measure of the shorter side = 4.5 cm

**Question 5. In a parallelogram ABCD, âˆ D = 135**Â°**. Determine the measures of âˆ A and âˆ B.**

**Solution:**

In a parallelogram ABCD

Given: âˆ D=135Â°

So, âˆ D + âˆ C = 180Â° [Consecutive angles of a parallelogram are supplementary]

âˆ C = 180Â°âˆ’ 135Â°

âˆ C = 45Â°

In a parallelogram opposite sides are equal.

âˆ A = âˆ C = 45Â° [opposite sides of parallelogram are equal]

âˆ B = âˆ D = 135Â°

hence, the measures of âˆ A and âˆ B are 45Â°,135Â°respectively.

**Question 6. ABCD is a parallelogram in which âˆ A = 70**Â°**. Compute âˆ B, âˆ C and âˆ D.**

**Solution:**

In a parallelogram ABCD

Given: âˆ A = 70Â°

âˆ A + âˆ B = 180Â° [Consecutive angles of a parallelogram are supplementary]

70Â°+ âˆ B = 180Â° [given âˆ A = 70Â°]

âˆ B = 180Â°âˆ’ 70Â°

âˆ B = 110Â°

Now,

âˆ A = âˆ C = 70Â° [opposite sides of parallelogram are equal]

âˆ B = âˆ D = 110Â°

hence, the measures of âˆ A and âˆ B are 70Â°,110Â°respectively.

**Question 7. In Figure, ABCD is a parallelogram in which âˆ A = 60**Â°**. If the bisectors of âˆ A, and âˆ B meet at P, prove that AD = DP, PC = BC and DC = 2AD.**

**Solution:**

Given: âˆ A = 60Â°

To prove: AD = DP, PC = BC and DC = 2AD

AP bisects âˆ A

so, âˆ DAP = âˆ PAB = 30Â°

Now,

âˆ A + âˆ B = 180Â° [Consecutive angles of a parallelogram are supplementary]

âˆ B + 60Â°= 180Â°

âˆ B = 180Â°âˆ’ 60Â°

âˆ B = 120Â°

BP bisects âˆ B

so, âˆ PBA = âˆ PBC = 60Â°

âˆ PAB = âˆ APD = 30Â°[Alternate interior angles]

Therefore, AD = DP [Sides opposite to equal angles are in equal length]

Similarly

âˆ PBA = âˆ BPC = 60Â° [Alternate interior angles]

Therefore, PC = BC

DC = DP + PC

DC = AD + BC [ DP = AD and PC = BC ]

DC = 2AD [Since, AD = BC, The opposite sides of parallelogram are parallel and congruent]

hence proved.

**Question 8. In figure, ABCD is a parallelogram in which âˆ DAB = 75**Â°**and âˆ DBC = 60**Â°**. Compute âˆ CDB, and âˆ ADB.**

**Solution:**

Given: âˆ DAB = 75Â°and âˆ DBC = 60Â°

To find âˆ CDB and âˆ ADB

âˆ CBD = âˆ ADB = 60Â° [Alternate interior angle. ADâˆ¥ BC and BD is the transversal]

InBDA

âˆ DAB + âˆ ADB + âˆ ABD = 180Â° [Angle sum property]

75Â°+ 60Â°+ âˆ CDB = 180Â°

âˆ ABD = 180Â°âˆ’ (135Â°)

âˆ ABD = 45Â°

âˆ ABD = âˆ CDB = 45Â° [Alternate interior angle. ADâˆ¥ BC and BD is the transversal]

Hence, âˆ CDB = 45Â°, âˆ ADB = 60Â°

**Question 9. In figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.**

**Solution:**

Given: ABCD is a parallelogram and E is the mid-point of side BC.

To prove: AF = 2AB.

Now,

In Î”BEF and Î”CED

âˆ BEF = âˆ CED [Verified opposite angle]

BE = CE [Since, E is the mid-point of BC]

âˆ EBF = âˆ ECD [Since, Alternate interior angles are equal]

Î”BEF â‰… Î”CED [ASA congruence]

BF = CD [Corresponding Parts of Congruent Triangle]

AF = AB + AF

AF = AB + CD [BF = CD by Corresponding Parts of Congruent Triangle ]

AF = AB + AB [CD=AB, The opposite sides of parallelogram are parallel and congruent]

AF = 2AB.

Hence proved.

**Question 10. 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

(ii) True

(iii) False

(iv) False

(v) True

(vi) False

(vii) False

(viii) True

## Please

Loginto comment...