The document RD Sharma Solutions Ex-14.1, Quadrilaterals, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**1) Three angles of a quadrilateral are respectively equal to 110 ^{0}, 50^{0} and 40^{0}. Find its fourth angle.**

**Solution:**

Given,

Three angles are 110^{0}, 50^{0} and 40^{0}

Let the fourth angle be ‘x’

We have,

Sum of all angles of a quadrilateral = 360^{0}

110^{0} + 50^{0} + 40^{0} = 360^{0}

⇒ x = 360^{0} – 200^{0}

⇒x = 160^{0}

Therefore, the required fourth angle is 160^{0}.

*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^{0}

⇒ x + 2x + 4x + 5x = 360^{0}

⇒ 12x = 360^{0}

⇒

⇒ x = 30^{0}

Therefore, A = x = 30^{0}

B = 2x = 60^{0}

C = 4x = 120^{0}

D = 5x = 150^{0}

**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=180^{0} [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=360^{0}−(∠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 360^{0}

Therefore, 3t + 5t + 9t + 13t = 360^{0}

⇒ 30t = 360^{0}

⇒ t = 12^{0}

Hence, the angles are

3t = 3*12 = 36^{0}

5t = 5*12 = 60^{0}

9t = 9*12 = 108^{0}

13t = 13*12 = 156^{0}

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