# Class 8 RD Sharma Solutions- Chapter 16 Understanding Shapes Quadrilaterals – Exercise 16.1 | Set 2

Last Updated : 08 Apr, 2021

### Question 13. In Figure, find the measure of âˆ MPN.

Solution:

As we know that Sum of angles of a quadrilateral is = 360Â°

âˆ NOP = 45Â°, âˆ OMP = âˆ PNO = 90Â° (Given)

Let us assume that angle âˆ MPN is xÂ°

âˆ NOP + âˆ OMP + âˆ PNO + âˆ MPN = 360Â°

45Â° + 90Â° + 90Â° + xÂ° = 360Â°

xÂ° = 360Â° â€“ 225Â°

xÂ° = 135Â°

Hence, Measure of âˆ MPN is 135Â°

### Question 14. The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles?

Solution:

As we know that, exterior angle + interior adjacent angle = 180Â° (Linear pair)

Applying relation for polygon having n sides

Sum of all exterior angles + Sum of all interior angles = n Ã— 180Â°

Sum of all exterior angles = n Ã— 180Â° â€“ Sum of all interior angles

= n Ã— 180Â° â€“ (n -2) Ã— 180Â° (Sum of interior angles is = (n â€“ 2) x 180Â°)

= n Ã— 180Â° â€“ n Ã— 180Â° + 2 Ã— 180Â°

= 180Â°n â€“ 180Â°n + 360Â° = 360Â°

Hence, Sum of four exterior angles is 360o

### Question 15. In Figure, the bisectors of âˆ A and âˆ B meet at a point P. If âˆ C =100Â° and âˆ D = 50Â°, find the measure of âˆ APB.

Solution:

As we know that Sum of angles of a quadrilateral is = 360Â°

Given that,

âˆ C =100Â° and âˆ D = 50Â°

âˆ A + âˆ B + âˆ C + âˆ D = 360o

âˆ A + âˆ B + 100o + 50o = 360o

âˆ A + âˆ B = 360o â€“ 150o

âˆ A + âˆ B = 210o (Equation 1)

Now in Î” APB

Â½ âˆ A + Â½ âˆ B + âˆ APB = 180o (sum of triangle is 180o)

âˆ APB = 180o â€“ Â½ (âˆ A + âˆ B) (Equation 2)

On substituting value of âˆ A + âˆ B = 210 from equation (1) in equation (2)

âˆ APB = 180o â€“ Â½ (210o)

= 180o â€“ 105o = 75o

Hence, the measure of âˆ APB is 75o.

### Question 16. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1:2:4:5. Find the measure of each angle of the quadrilateral.

Solution:

As we know that Sum of angles of a quadrilateral is = 360Â°

Let each angle be xo

Therefore,

xo + 2xo + 4xo + 5xo = 360o

12xo = 360o

xo = 360o/12 = 30o

Value of angles are as x = 30o,

2x = 2 Ã— 30 = 60o

4x = 4 Ã— 30 = 120o

5x = 5 Ã— 30 = 150o

Hence, Value of angles are 30o, 60o, 120o, 150o.

### Question 17. 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 +âˆ B).

Solution:

As we know that sum of angles of a quadrilateral is 360Â°

Therefore,

âˆ A + âˆ B + âˆ C + âˆ D = 360o

âˆ A + âˆ B = 360o â€“ (âˆ C + âˆ D)

Â½ (âˆ A + âˆ B) = Â½ [360o â€“ (âˆ C + âˆ D)]

= 180o â€“ Â½ (âˆ C + âˆ D) (Equation 1)

Now in Î” DOC

Â½ âˆ D + Â½ âˆ C + âˆ COD = 180o (We know that sum of triangle = 180o)

Â½ (âˆ C + âˆ D) + âˆ COD = 180o

âˆ COD = 180o â€“ Â½ (âˆ C + âˆ D) (Equation 2)

In equations (1) and (2) RHS is equal then LHS will also equal.

Hence, âˆ COD = Â½ (âˆ A + âˆ B) is proved.

### Question 18. Find the number of sides of a regular polygon, when each of its angles has a measure of

(i) 160Â°

(ii) 135Â°

(iii) 175Â°

(iv) 162Â°

(v) 150Â°

Solution:

The sum of interior angle A of a polygon of n sides is given by A = [(n-2) Ã—180o] /n

(i) 160o

Angle of quadrilateral is 160Â° (Given)

160o = [(n-2) Ã—180o]/n

160on = (n-2) Ã—180o

160on = 180on â€“ 360o

180on â€“ 160on = 360o

20on = 360o

n = 360o/20 = 18

Hence Number of sides are 18

(ii) 135o

Angle of quadrilateral is 135Â° (Given)

135o = [(n-2) Ã—180o]/n

135on = (n-2) Ã—180o

135on = 180on â€“ 360o

180on â€“ 135on = 360o

45on = 360o

n = 360o / 45 = 8

Hence Number of sides are 8

(iii) 175o

Angle of quadrilateral is 175Â° (Given)

175o = [(n-2) Ã—180o]/n

175on = (n-2) Ã—180o

175on = 180on â€“ 360o

180on â€“ 175on = 360o

5on = 360o

n = 360o/5 = 72

Hence Number of sides are 72

(iv) 162o

Angle of quadrilateral is 162Â° (Given)

162o = [(n-2) Ã—180o]/n

162on = (n-2) Ã—180o

162on = 180on â€“ 360o

180on â€“ 162on = 360o

18on = 360o

n = 360o/18 = 20

Hence Number of sides are 20

(v) 150o

Angle of quadrilateral is 160Â° (Given)

150o = [(n-2) Ã—180o]/n

150on = (n-2) Ã—180o

150on = 180on â€“ 360o

180on â€“ 150on = 360o

30on = 360o

n = 360o/30 = 12

Hence Number of sides are 12

### Question 19. Find the numbers of degrees in each exterior angle of a regular pentagon.

Solution:

As we know that the sum of exterior angles of a polygon is 360Â°

Sum of each exterior angle of a polygon = 360o/n (n is the number of sides)

As we know that number of sides in a pentagon is 5

Sum of each exterior angle of a pentagon = 360o/5 = 72o

Hence Measure of each exterior angle of a pentagon is 72o

### Question 20. The measure of angles of a hexagon are xÂ°, (x-5)Â°, (x-5)Â°, (2x-5)Â°, (2x-5)Â°, (2x+20)Â°. Find value of x.

Solution:

As we know that the sum of interior angles of a polygon = (n â€“ 2) Ã— 180Â° (n = number of sides of polygon)

As we know that hexagon has 6 sides therefore,

The sum of interior angles of a hexagon = (6 â€“ 2) Ã— 180Â° = 4 Ã— 180Â° = 720Â°

xÂ°+ (x-5)Â°+ (x-5)Â°+ (2x-5)Â°+ (2x-5)Â°+ (2x+20)Â° = 720Â°

xÂ°+ xÂ°- 5Â°+ xÂ° â€“ 5Â°+ 2xÂ° â€“ 5Â°+ 2xÂ° â€“ 5Â°+ 2xÂ° + 20Â° = 720Â°

9xÂ° = 720Â°

x = 720o/9 = 80o

Hence Value of x is 80o

### Question 21. In a convex hexagon, prove that the sum of all interior angle is equal to twice the sum of its exterior angles formed by producing the sides in the same order.

Solution:

As we know that the sum of interior angles of a polygon = (n â€“ 2) Ã— 180Â°

The sum of interior angles of a hexagon = (6 â€“ 2) Ã— 180Â° = 4 Ã— 180Â° = 720Â°

Sum of exterior angle of a polygon is 360Â°

Hence Sum of interior angles of a hexagon = Twice the sum of interior angles.

Hence proved.

### Question 22. The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sides of the polygon.

Solution:

As we know that the sum of interior angles of a polygon = (n â€“ 2) Ã— 180Â° (i)

The Sum of exterior angle of a polygon is 360Â°

therefore,

Sum of Interior Angles = 3 Ã— sum of exterior angles

= 3 Ã— 360Â° = 1080Â° (ii)

Now by equating (i) and (ii) we get,

(n â€“ 2) Ã— 180Â° = 1080Â°

n â€“ 2 = 1080o/180o

n â€“ 2 = 6

n = 6 + 2 = 8

Hence Number of sides of a polygon is 8.

### Question 23. Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5.

Solution:

As we know that the sum of interior angles of a polygon = (n â€“ 2) Ã— 180Â° (i)

The Sum of exterior angle of a polygon is 360Â°

As we know that Sum of exterior angles / Sum of interior angles = 1/5 (ii)

By equating (i) and (ii) we get,

360o/(n â€“ 2) Ã— 180Â° = 1/5

(n â€“ 2) Ã— 180Â° = 360o Ã— 5

(n â€“ 2) Ã— 180Â° = 1800o

(n â€“ 2) = 1800o/180o

(n â€“ 2) = 10

n = 10 + 2 = 12

Hence Numbers of sides of a polygon is 12.

### Question 24. PQRSTU is a regular hexagon, determine each angle of Î”PQT.

Solution:

As we know that the sum of interior angles of a polygon = (n â€“ 2) Ã— 180Â°

The sum of interior angles of a hexagon = (6 â€“ 2) Ã— 180Â° = 4 Ã— 180Â° = 720Â°

Sum of each angle of hexagon = 720o/6 = 120o

âˆ PUT = 120o Proved.

In Î” PUT

âˆ PUT + âˆ UTP + âˆ TPU = 180o (sum of triangles)

120o + 2âˆ UTP = 180o (Since Î” PUT is an isosceles triangle )

2âˆ UTP = 180o â€“ 120o

2âˆ UTP = 60o

âˆ UTP = 60o/2 = 30o

âˆ UTP = âˆ TPU = 30o similarly âˆ RTS = 30o

therefore âˆ PTR = âˆ UTS â€“ âˆ UTP â€“ âˆ RTS

= 120o â€“ 30o â€“ 30o = 60o

âˆ TPQ = âˆ UPQ â€“ âˆ UPT

= 120o â€“ 30o = 90o

âˆ TQP = 180o â€“ 150o = 30o (By using angle sum property of triangle in Î”PQT)

Hence âˆ P = 90o, âˆ Q = 60o, âˆ T = 30o

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