Chapter 14 Quadrilaterals RD Sharma Solutions Exercise 14.2 Class 9 Maths
Chapter Name  RD Sharma Chapter 14 Quadrilaterals Exercise 14.2 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 14.2 Solutions
1. Two opposite angles of a parallelogram are (3x  2)° and (50  x)° . Find the measure of each angle of the parallelogram.
Solution
We know that
Opposite sides of a parallelogram are equal
∴ 3x 2 = 50  x
⇒ 3x + x = 50 + 2
⇒ 4x = 52
⇒ x = 13°
∴ (3x  2)° = (3 × 13  2) = 37°
(50  x)° = (50  13°) = 37°
Adjacent angles of a parallelogram are supplementary
∴ x + 37 = 180°
∴ x = 180°  37° = 143°
Hence, four angles are : 37°, 143°, 37° , 143°
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
∴ The measure of the angle adjacent is 2x/3
We know that the adjacent angle of a parallelogram is supplementary
Hence x + 2x/3 = 180°
2x + 3x = 540°
⇒ 5x = 540°
⇒ x = 108°
Adjacent angles are supplementary
⇒ x + 108° = 180°
⇒ x = 180°  108° = 72°
⇒ x = 72°
Hence, four angles are : 180° , 72°, 108°, 72°
3. Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.
Solution
Let the smallest angle be x
Then, the other angle is (3x  24)
Now, x + 2x  24 = 180°
3x  24 = 180°
⇒ 3x = 180° + 24
⇒ 3x = 204°
⇒ x = 204/3 = 68°
⇒ x = 68°
⇒ 2x 24° = 2×68°  24° = 136°  24° = 112°
Hence four angles are 68°, 112°, 68°, 112°
4. The perimeter of a parallelogram is 22cm. If the longer side measures 6.5 cm what is the measure of the shorter side ?
Solution
Let the shorter side be x
∴Perimeter = x + 6.5 + 6.5 + x
22 = 2(x + 6.5)
11 = x + 6.5
⇒ x = 11  6.5 = 4.5 cm
∴ Shorter side = 4.5 cm
5. In a parallelogram ABCD, ∠D = 135°, determine the measures of ∠A and ∠B.
Solution
In a parallelogram ABCD
Adjacent angles are supplementary
So, ∠D + ∠C = 180°
135 + ∠C = 180°
⇒ ∠C = 180°  135°
⇒ ∠C = 45°
In a parallelogram opposite sides are equal
∠A = ∠C = 45°
∠B = ∠D = 135°
6. ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.
Solution
In a parallelogram ABCD
∠A = 70° [∵ Adjacent angles supplementary]
∠A = ∠B = 180°
⇒ 70° + ∠B = 180° [∵ ∠A = 70°]
⇒ ∠B = 180°  70° = 110°
In a parallelogram opposite sides are equal
∠A = ∠C = 70°
∠B = ∠D = 110°
7. In fig., below, 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
AP bisects ∠AThen, ∠AP = ∠PAB = 30°
Adjacent angles are supplementary
BP bisects ∠B
Then, ∠PBA + ∠PBC = 30°
Similarly
∠PBA = ∠BPC = 60° [Alternative interior angle]
DC = DP + PC
⇒ DC = AD + BC [∵ DP = AD, PC = BC]
⇒ DC = 2AD
∠BEF = ∠CED [Verified opposite angle]
BE = CE [∵ E is the mid  point of BC]
∵ Î”BEF ≅ Î”CED [Angle side angle congruence]
∵ BF = CD [Corresponding Parts of Congruent Triangles]
AF = AB + AB
AF = 2AB