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

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

Solution:

Given,
The three angles are 110Â°, 50Â° and 40Â°
Let the fourth angle be ‘x’
Sum of all angles of a quadrilateral = 360Â°
110Â° + 50Â° + 40Â° + x = 360Â°
200Â° + x = 360Â°
x = 360Â° âˆ’ 200Â°
x = 160Â°
Hence, the required fourth angle is 160Â°.

### 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 = 360Â° {Sum of interior angle of quadrilateral 360Â°}
x + 2x + 4x + 5x = 360Â°
12x = 360Â°
x = 360Â°
12
x = 30Â°
Therefore, substituting value of x,
A = x = 30Â°
B = 2x = 2 Ã— 30Â° = 60Â°
C = 4x = 4 Ã— 30Â° = 120Â°
D = 5x = 5 Ã— 30Â° = 150Â°

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

Solution:

In Î”DOC,
âˆ CDO + âˆ COD + âˆ DCO = 1800 [Angle sum property of a triangle]
or
1 âˆ CDA + âˆ COD +  1 âˆ DCB = 1800
2                               2
âˆ COD = 1800 â€“  1 (âˆ CDA + âˆ DCB) —->(equation 1)
2
Also
We know, sum of all angles of a quadrilateral = 360Â°
âˆ CDA + âˆ DCB = 3600 â€“ (âˆ DAB + âˆ CBA) —->(equation 2)
Substituting equation 1 and equation 2
âˆ COD = 1800 â€“  1 {3600 â€“ (âˆ DAB + âˆ CBA)}
2
We can also write,
âˆ DAB = âˆ A and âˆ CBA = âˆ B
âˆ COD = 1800 âˆ’ 1800 + 1/2(âˆ A + âˆ B))
âˆ COD = 1/2(âˆ A + âˆ 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 = 360Â° {{Sum of interior angle of quadrilateral 360Â°}
3x + 5x + 9x + 13x = 360Â°
30x = 360Â°
x =  360Â°
30
x = 12Â°
Therefore, substituting value of x,
A = 3x = 3Ã—12 = 36Â°
B = 5x = 5Ã—12Â° = 60Â°
C = 9x = 9Ã—12Â° = 108Â°
D = 13x = 13Ã—12Â° = 156Â°

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