RD Sharma Solutions Ex-14.1, Quadrilaterals, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

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 = 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 ∠Cand∠D respectively. Prove that ∠COD = 1/2(∠Aand∠B).

Solution:

InΔDOC

∠1+∠COD+∠2=180⁰          [Angle sum property of a triangle]

⇒ ∠COD = 180−(∠1−∠2)

⇒ ∠COD = 180−∠1+∠2

⇒   [∵ OC and Od are bisectors of LC and LD respectively]

⇒ 

In quadrilateral ABCD

∠A+∠B+∠C+∠D = 3600  [Angle sum propert yof quadrilateral]

∠C+∠D=360⁰−(∠A+∠B)….(ii)

Substituting (ii) in (i)

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