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,

Three angles are 110^°, 50^° and 40^°

Let the fourth angle be 'x'

We have,

Sum of all angles of a quadrilateral = 360^°

110^° + 50^° + 40^° = 360^°

⟹ x = 360^° - 200^°

⟹ x = 160^°

Therefore, 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 the quadrilaterals be

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

Then,

A + B + C + D = 360^°

⟹ x + 2x + 4x + 5x = 360^°

⟹ 12x = 360^°

⟹ x = 360°/12

⟹ x = 30^°

Therefore, A = x = 30^°

B = 2x = 60^°

C = 4x = 120^°

D = 5x = 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).

Solution:

In ΔDOC

∠1 + ∠COD + ∠2 = 180°      [Angle sum property of a triangle]

⟹ ∠COD = 180 − (∠1 − ∠2)

⟹ ∠COD = 180 − ∠1 + ∠2

⟹ ∠COD = 180 − [1/2 LC + 1/2 LD]        [∵ OC and OD are bisectors of LC and LD respectively]

⟹ ∠COD = 180 – 1/2(LC + LD)      ... (i)

In quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360°  [Angle sum property of quadrilateral]

∠C + ∠D = 360° − (∠A + ∠B)   .... (ii)

Substituting (ii) in (i)

⟹ ∠COD = 180 – 1/2(360 − (∠A + ∠B))

⟹ ∠COD = 180 − 180 +1/2(∠A + ∠B))

⟹ ∠COD = 1/2(∠A + ∠B))

 

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 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 360^°

Therefore, 3t + 5t + 9t + 13t = 360^°

⟹ 30t = 360^°

⟹ t = 12^°

Hence, the angles are

3t = 3*12 = 36^°

5t = 5*12 = 60^°

9t = 9*12 = 108^°

13t = 13*12 = 156^°