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

### Chapter 16 Understanding Shapes Quadrilaterals – Exercise 16.1 | Set 1

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

**Solution:**

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

In the quadrilateral MPNO

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 360^{o}

**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°

In the quadrilateral ABCD

Given that,

âˆ C =100Â° and âˆ D = 50Â°

âˆ A + âˆ B + âˆ C + âˆ D = 360

^{o}âˆ A + âˆ B + 100

^{o}+ 50^{o}= 360^{o}âˆ A + âˆ B = 360

^{o}â€“ 150^{o}âˆ A + âˆ B = 210

^{o}(Equation 1)Now in Î” APB

Â½ âˆ A + Â½ âˆ B + âˆ APB = 180

^{o}(sum of triangle is 180^{o})âˆ APB = 180

^{o}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 = 180

^{o}â€“ Â½ (210^{o})= 180

^{o}â€“ 105^{o}= 75^{o}

Hence, the measure of âˆ APB is 75^{o}.

**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 x

^{o}Therefore,

x

^{o}+ 2x^{o}+ 4x^{o}+ 5x^{o}= 360^{o}12x

^{o}= 360^{o}x

^{o}= 360^{o}/12 = 30^{o}Value of angles are as x = 30

^{o},2x = 2 Ã— 30 = 60

^{o}4x = 4 Ã— 30 = 120

^{o}5x = 5 Ã— 30 = 150

^{o}

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

**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°

In the quadrilateral ABCD

Therefore,

âˆ A + âˆ B + âˆ C + âˆ D = 360

^{o}âˆ A + âˆ B = 360o â€“ (âˆ C + âˆ D)

Â½ (âˆ A + âˆ B) = Â½ [360

^{o}â€“ (âˆ C + âˆ D)]= 180

^{o}â€“ Â½ (âˆ C + âˆ D)(Equation 1)Now in Î” DOC

Â½ âˆ D + Â½ âˆ C + âˆ COD = 180

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

^{o}âˆ COD = 180

^{o}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) 160^{o}Angle of quadrilateral is 160Â°

(Given)160

^{o}= [(n-2) Ã—180^{o}]/n160

^{o}n = (n-2) Ã—180^{o}160

^{o}n = 180^{o}n â€“ 360^{o}180

^{o}n â€“ 160^{o}n = 360^{o}20

^{o}n = 360^{o}n = 360

^{o}/20 = 18

Hence Number of sides are 18

(ii)135^{o}Angle of quadrilateral is 135Â°

(Given)135

^{o}= [(n-2) Ã—180^{o}]/n135

^{o}n = (n-2) Ã—180^{o}135

^{o}n = 180^{o}n â€“ 360^{o}180

^{o}n â€“ 135^{o}n = 360^{o}45

^{o}n = 360^{o}n = 360

^{o}/ 45 = 8

Hence Number of sides are 8

(iii) 175^{o}Angle of quadrilateral is 175Â°

(Given)175

^{o}= [(n-2) Ã—180^{o}]/n175

^{o}n = (n-2) Ã—180^{o}175

^{o}n = 180^{o}n â€“ 360^{o}180

^{o}n â€“ 175^{o}n = 360^{o}5

^{o}n = 360^{o}n = 360

^{o}/5 = 72

Hence Number of sides are 72

(iv)162^{o}Angle of quadrilateral is 162Â°

(Given)162

^{o}= [(n-2) Ã—180^{o}]/n162

^{o}n = (n-2) Ã—180^{o}162

^{o}n = 180^{o}n â€“ 360^{o}180

^{o}n â€“ 162^{o}n = 360^{o}18

^{o}n = 360^{o}n = 360

^{o}/18 = 20

Hence Number of sides are 20

(v) 150^{o}Angle of quadrilateral is 160Â°

(Given)150

^{o}= [(n-2) Ã—180^{o}]/n150

^{o}n = (n-2) Ã—180^{o}150

^{o}n = 180^{o}n â€“ 360^{o}180

^{o}n â€“ 150^{o}n = 360^{o}30

^{o}n = 360^{o}n = 360

^{o}/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 = 360

^{o}/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 = 360

^{o}/5 = 72^{o}

Hence Measure of each exterior angle of a pentagon is 72^{o}

**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 = 720

^{o}/9 = 80^{o}

Hence Value of x is 80^{o}

**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 = 1080

^{o}/180^{o}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,

360

^{o}/(n â€“ 2) Ã— 180Â° = 1/5(n â€“ 2) Ã— 180Â° = 360

^{o}Ã— 5(n â€“ 2) Ã— 180Â° = 1800

^{o}(n â€“ 2) = 1800

^{o}/180^{o}(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 = 720

^{o}/6 = 120^{o}âˆ PUT = 120

^{o}Proved.In Î” PUT

âˆ PUT + âˆ UTP + âˆ TPU = 180

^{o}(sum of triangles)120o + 2âˆ UTP = 180

^{o}(Since Î” PUT is an isosceles triangle )2âˆ UTP = 180

^{o}â€“ 120^{o}2âˆ UTP = 60

^{o}âˆ UTP = 60

^{o}/2 = 30^{o}âˆ UTP = âˆ TPU = 30

^{o}similarly âˆ RTS = 30^{o}therefore âˆ PTR = âˆ UTS â€“ âˆ UTP â€“ âˆ RTS

= 120

^{o}â€“ 30^{o}â€“ 30^{o}= 60^{o}âˆ TPQ = âˆ UPQ â€“ âˆ UPT

= 120

^{o}â€“ 30^{o}= 90^{o}âˆ TQP = 180

^{o}â€“ 150^{o}= 30^{o}(By using angle sum property of triangle in Î”PQT)

Hence âˆ P = 90^{o}, âˆ Q = 60^{o}, âˆ T = 30^{o}

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