# Class 9 RD Sharma Solutions – Chapter 14 Quadrilaterals- Exercise 14.1

**Question 1. Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.**

**Solution:**

Given,

The three angles are 110°, 50° and 40°

Let the fourth angle be ‘x’

Sum of all angles of a quadrilateral = 360°

110° + 50° + 40° + x = 360°

200° + x = 360°

x = 360° − 200°

x = 160°

Hence, the required fourth angle is 160°.

**Question 2. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio of 1:2:4:5. Find the measure of each angles of the quadrilateral.**

**Solution: **

Let the angles of quadrilateral be

A = x, B = 2x, C = 4x, D = 5x

Then,

A + B + C + D = 360° {Sum of interior angle of quadrilateral 360°}

x + 2x + 4x + 5x = 360°

12x = 360°

x =360°

12

x = 30°

Therefore, substituting value of x,

A = x = 30°

B = 2x = 2 × 30° = 60°

C = 4x = 4 × 30° = 120°

D = 5x = 5 × 30° = 150°

**Question 3. In a quadrilateral ABCD, CO and Do are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 1/2 **__(__∠A and ∠B)

__(__∠A and ∠B)

**Solution:**

In ΔDOC,

∠CDO + ∠COD + ∠DCO = 1800 [Angle sum property of a triangle]

or1∠CDA + ∠COD +1∠DCB = 1800

2_{ }2

∠COD = 1800 –1(∠CDA + ∠DCB) —->(equation 1)

2

Also

We know, sum of all angles of a quadrilateral = 360°

∠CDA + ∠DCB = 3600 – (∠DAB + ∠CBA) —->(equation 2)

Substituting equation 1 and equation 2

∠COD = 1800 –1{3600 – (∠DAB + ∠CBA)}

2

We can also write,

∠DAB = ∠A and ∠CBA = ∠B

∠COD = 1800 − 1800 + 1/2(∠A + ∠B))

∠COD = 1/2(∠A + ∠B)

Hence, Proved.

**Question 4. The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.**

**Solution:**

Let the angles of quadrilateral be

A = 3x, B = 5x, C = 9x, D = 13x

Then,

A + B + C + D = 360° {{Sum of interior angle of quadrilateral 360°}

3x + 5x + 9x + 13x = 360°

30x = 360°

x =360°

30

x = 12°

Therefore, substituting value of x,

A = 3x = 3×12 = 36°

B = 5x = 5×12° = 60°

C = 9x = 9×12° = 108°

D = 13x = 13×12° = 156°

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