# 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 ∠MPN.**

**Solution:**

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

In the quadrilateral MPNO

∠NOP = 45°, ∠OMP = ∠PNO = 90°

(Given)Let us assume that angle ∠MPN is x°

∠NOP + ∠OMP + ∠PNO + ∠MPN = 360°

45° + 90° + 90° + x° = 360°

x° = 360° – 225°

x° =

135°

Hence, Measure of ∠MPN is 135°

**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 = 180°

(Linear pair)Applying relation for polygon having n sides

Sum of all exterior angles + Sum of all interior angles = n × 180°

Sum of all exterior angles = n × 180° – Sum of all interior angles

= n × 180° – (n -2) × 180°

(Sum of interior angles is = (n – 2) x 180°)= n × 180° – n × 180° + 2 × 180°

= 180°n – 180°n + 360° =

360°

Hence, Sum of four exterior angles is 360^{o}

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

**Solution:**

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

In the quadrilateral ABCD

Given that,

∠C =100° and ∠D = 50°

∠A + ∠B + ∠C + ∠D = 360

^{o}∠A + ∠B + 100

^{o}+ 50^{o}= 360^{o}∠A + ∠B = 360

^{o}– 150^{o}∠A + ∠B = 210

^{o}(Equation 1)Now in Δ APB

½ ∠A + ½ ∠B + ∠APB = 180

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

^{o}– ½ (∠A + ∠B)(Equation 2)On substituting value of ∠A + ∠B = 210 from equation (1) in equation (2)

∠APB = 180

^{o}– ½ (210^{o})= 180

^{o}– 105^{o}= 75^{o}

Hence, the measure of ∠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 = 360°

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 × 30 = 60

^{o}4x = 4 × 30 = 120

^{o}5x = 5 × 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 ∠C and ∠D respectively. Prove that ∠COD = 1/2 (∠A +∠B).**

**Solution:**

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

In the quadrilateral ABCD

Therefore,

∠A + ∠B + ∠C + ∠D = 360

^{o}∠A + ∠B = 360o – (∠C + ∠D)

½ (∠A + ∠B) = ½ [360

^{o}– (∠C + ∠D)]= 180

^{o}– ½ (∠C + ∠D)(Equation 1)Now in Δ DOC

½ ∠D + ½ ∠C + ∠COD = 180

^{o}(We know that sum of triangle = 180o)½ (∠C + ∠D) + ∠COD = 180

^{o}∠COD = 180

^{o}– ½ (∠C + ∠D)(Equation 2)In equations (1) and (2) RHS is equal then LHS will also equal.

Hence, ∠COD = ½ (∠A + ∠B) is proved.

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

**(i) 160°**

**(ii) 135°**

**(iii) 175°**

**(iv) 162°**

**(v) 150°**

**Solution:**

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

(i) 160^{o}Angle of quadrilateral is 160°

(Given)160

^{o}= [(n-2) ×180^{o}]/n160

^{o}n = (n-2) ×180^{o}160

^{o}n = 180^{o}n – 360^{o}180

^{o}n – 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 135°

(Given)135

^{o}= [(n-2) ×180^{o}]/n135

^{o}n = (n-2) ×180^{o}135

^{o}n = 180^{o}n – 360^{o}180

^{o}n – 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 175°

(Given)175

^{o}= [(n-2) ×180^{o}]/n175

^{o}n = (n-2) ×180^{o}175

^{o}n = 180^{o}n – 360^{o}180

^{o}n – 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 162°

(Given)162

^{o}= [(n-2) ×180^{o}]/n162

^{o}n = (n-2) ×180^{o}162

^{o}n = 180^{o}n – 360^{o}180

^{o}n – 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 160°

(Given)150

^{o}= [(n-2) ×180^{o}]/n150

^{o}n = (n-2) ×180^{o}150

^{o}n = 180^{o}n – 360^{o}180

^{o}n – 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 360°

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 x°, (x-5)°, (x-5)°, (2x-5)°, (2x-5)°, (2x+20)°. Find value of x.**

**Solution:**

As we know that the sum of interior angles of a polygon = (n – 2) × 180°

(n = number of sides of polygon)As we know that hexagon has 6 sides therefore,

The sum of interior angles of a hexagon = (6 – 2) × 180° = 4 × 180° = 720°

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

x°+ x°- 5°+ x° – 5°+ 2x° – 5°+ 2x° – 5°+ 2x° + 20° = 720°

9x° = 720°

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 – 2) × 180°

The sum of interior angles of a hexagon = (6 – 2) × 180° = 4 × 180° = 720°

Sum of exterior angle of a polygon is 360°

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 – 2) × 180°

(i)The Sum of exterior angle of a polygon is 360°

therefore,

Sum of Interior Angles = 3 × sum of exterior angles

= 3 × 360° = 1080°

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

(n – 2) × 180° = 1080°

n – 2 = 1080

^{o}/180^{o}n – 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 – 2) × 180°

(i)The Sum of exterior angle of a polygon is 360°

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 – 2) × 180° = 1/5(n – 2) × 180° = 360

^{o}× 5(n – 2) × 180° = 1800

^{o}(n – 2) = 1800

^{o}/180^{o}(n – 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 ΔPQT.**

**Solution: **

As we know that the sum of interior angles of a polygon = (n – 2) × 180°

The sum of interior angles of a hexagon = (6 – 2) × 180° = 4 × 180° = 720°

Sum of each angle of hexagon = 720

^{o}/6 = 120^{o}∠PUT = 120

^{o}Proved.In Δ PUT

∠PUT + ∠UTP + ∠TPU = 180

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

^{o}(Since Δ PUT is an isosceles triangle )2∠UTP = 180

^{o}– 120^{o}2∠UTP = 60

^{o}∠UTP = 60

^{o}/2 = 30^{o}∠UTP = ∠TPU = 30

^{o}similarly ∠RTS = 30^{o}therefore ∠PTR = ∠UTS – ∠UTP – ∠RTS

= 120

^{o}– 30^{o}– 30^{o}= 60^{o}∠TPQ = ∠UPQ – ∠UPT

= 120

^{o}– 30^{o}= 90^{o}∠TQP = 180

^{o}– 150^{o}= 30^{o}(By using angle sum property of triangle in ΔPQT)

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

Attention reader! Don’t stop learning now. Participate in the **Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students**.