# RD Sharma Solutions Class 9 Quadrilaterals Exercise 14.1

## RD Sharma Solutions Class 9 Chapter 14 Ex 14.1

1) Three angles of a quadrilateral are respectively equal to 1100, 500 and 400. Find its fourth angle.

Solution:

Given,

Three angles are 1100, 500 and 400

Let the fourth angle be ‘x’

We have,

Sum of all angles of a quadrilateral = 3600

1100 + 500 + 400 = 3600

=> x = 3600 – 2000

=>x = 1600

Therefore, the required fourth angle is 1600.

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 the quadrilaterals be

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

Then,

A + B + C + D = 3600

=> x + 2x + 4x + 5x = 3600

=> 12x = 3600

=> $x=\frac{360^{0}}{12}$

=> x = 300

Therefore, A = x = 300

B = 2x = 600

C = 4x = 1200

D = 5x = 1500

3) In a quadrilateral ABCD, CO and Do are the bisectors of $\angle C\;and\angle D$ respectively. Prove that $\angle COD=\frac{1}{2}(\angle A\;and\angle B)$.

Solution:

$In\;\Delta DOC$

$\angle 1+\angle COD+\angle 2=180^{0}$          [Angle sum property of a triangle]

=> $\angle COD=180-(\angle 1-\angle 2)$

=> $\angle COD=180-\angle 1+\angle 2$

=> $\angle COD=180-[\frac{1}{2}LC+\frac{1}{2}LD]\;\;\;[∵ OC\;and\;Od\;are\;bisectors\;of\;LC\;and\;LD\;respectively]$

=> $\angle COD=180-\frac{1}{2}(LC+LD)….(i)$

$\angle A+\angle B+\angle C+\angle D=360^{0}\;\;\;\;\;\;[Angle\;sum\;property\;of\;quadrilateral]$

$\angle C+\angle D=360^{0}-(\angle A+\angle B)$….(ii)

Substituting (ii) in (i)

=>$\angle COD=180-\frac{1}{2}(360-(\angle A+\angle B))$

=>$\angle COD=180-180+\frac{1}{2}(\angle A+\angle B))$

=>$\angle COD=\frac{1}{2}(\angle A+\angle B))$

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

Solution:

Let the common ratio between the angles is ‘t’

So the angles will be 3t, 5t, 9t and 13t respectively.

Since the sum of all interior angles of a quadrilateral is 3600

Therefore, 3t + 5t + 9t + 13t = 3600

=>30t = 3600

=>t = 120

Hence, the angles are

3t = 3*12 = 360

5t = 5*12 = 600

9t = 9*12 = 1080

13t = 13*12 = 1560

#### Practise This Question

π20dxa2cos2x+b2sin2x where (a, b >0) is equal to-