# Selina Concise Solutions for Chapter 16 Understanding Shapes Class 8 ICSE Mathematics

**Exercise 16A**

**1. State which of the following are polygons:**

**If the given figure is a polygon, name it as convex or concave.**

**Solution**

Only Fig. (ii), (iii) and (v) are polygons.

Fig. (ii) and (iii) are concave polygons while

Fig. (v) is convex.

**2. Calculate the sum of angles of a polygon with :**

**(i) 10 sides**

**(ii) 12 sides**

**(iii) 20 sides**

**(iv) 25 sides**

**Solution**

(i) No. of sides n = 10

sum of angles of polygon = (n – 2)×180°

= (10 – 2)× 180° = 1440°

(ii) no. of sides n = 12

sum of angles = (n – 2)×180°

= (12 – 2)×180° = 10×180° = 1800°

(iii) n = 20

Sum of angles of Polygon = (n – 2)× 180°

= (20 – 2)× 180° = 3240°

(iv) n = 25

Sum of angles of polygon = (n – 2)× 180°

= (25 – 2)× 180° = 4140°

**3. Find the number of sides in a polygon if the sum of its interior angles is :**

**(i) 900°**

**(ii) 1620°**

**(iii) 16 right-angles**

**(iv) 32 right-angles.**

**Solution**

(i) Let no. of sides = n

Sum of angles of polygon = 900˚

(n – 2) × 180˚ = 900˚

⇒ n – 2 = 900/180

⇒ n – 2 = 5

⇒ n = 5 + 2

⇒ n = 7

(ii) Let no. of sides = n

Sum of angles of polygon = 1620˚

(n – 2) × 180˚ = 1620˚

⇒ n – 2 = 1620/180

⇒ n – 2 = 9

⇒ n = 9 + 2

⇒ n = 11

(iii) Let no. of sides = n

Sum of angles of polygon = 16 right = 16 × 90 = 1440˚

(n – 2) × 180˚ = 1440˚

⇒ n – 2 = 1440/180˚

⇒ n – 2 = 8

⇒ n = 8 + 2

⇒ n = 10

(iv) Let no. of sides = n

Sum of angles of polygon = 32 right angles = 32 × 90 = 2880˚

(n × 2) × 180˚ = 2880

n – 2 = 2880/180

n – 2 = 16

n = 16 + 2

n =18

(i) Let no. of sides = n

Sum of angles = 870°

(n – 2) × 180° = 870°

⇒ n – 2 = 870/180

⇒ n – 2 = 29/6

⇒ n = 29/6 + 2

⇒ n = 41/6

Which is not a whole number.

Hence it is not possible to have a polygon, the sum of whose interior angles is 870°

(ii) Let no. of sides = n

Sum of angles = 2340°

(n – 2) × 180° = 2340°

⇒ n – 2 = 2340/180

⇒ n – 2 = 13

⇒ n = 13 + 2 = 15

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.

(iii) Let no. of sides = n

Sum of angles = 7 right angles = 7 ×90 = 630°

(n – 2) × 180° = 630°

⇒ n – 2 = 630/180

⇒ n – 2 = 7/2

⇒ n = 7/2 + 2

⇒ n = 11/2

Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.

(iv) Let no. of sides = n

(n – 2)×180° = 4500°

⇒ n – 2 = 4500/180

⇒ n – 2 = 25

⇒ n = 25 + 2

⇒ n = 27

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.

(i) No. of sides of hexagon, n = 6

Let each angle be = x°

Sum of angles = 6x°

(n – 2)× 180° = Sum of angles

⇒ (6 – 2)× 180° = 6x°

⇒ 4× 180 = 6x

⇒ x = (4 × 180)/6

⇒ x = 120˚

∴ Each angle of hexagon = 120˚

(ii) No. of sides of polygon, n = 14

Let each angle = x˚

∴ Sum of angles = 14x˚

∴ (n – 2) × 180˚ = Sum of angles of polygon

∴ (14 – 2) × 180˚ = 14x

12 × 180˚ = 14x

⇒ x = (12 × 180)/14

⇒ x = 1080/7

⇒ x = (154 2/7)˚

(i) No. of sides n = 7

Sum of interior & exterior angles at one vertex = 180°

Sum of interior & exterior angles = 7 × 180˚

= 1260˚

Sum of interior angles = (n – 2) × 180˚

= (7 – 2) × 180˚

= 900˚

∴ Sum of exterior angles = 1260˚ - 900˚

= 360˚

(ii) No. of sides n = 10

Sum of interior and exterior angles = 10 × 180˚

= 1800˚

But sum of interior angles = (n – 2) × 180˚

= (10 – 2) × 180˚

= 1440˚

∴ Sum of exterior angles = 1800 – 1440

= 360˚

(iii) No. of sides n = 250

Sum of all interior and exterior angles = 250 × 180˚

= 45000˚

But sum of interior angles = (n – 2) × 180˚

= (250 – 2) × 180˚

= 248 × 180˚

= 44640˚

∴ Sum of exterior angles = 45000 – 44640

= 360˚

Sum of exterior angles of hexagon formed by producing sides of order = 360°

∴ (6x – 1)˚ - (10x + 2)˚ + (8x + 2)˚ + (9x – 3)˚ + (5x + 4)˚ + (12x + 6)˚ = 360˚

50x + 10˚ = 360˚

⇒ 50x = 360˚ - 10˚

⇒ 50x = 350˚

⇒ x = 350/70

⇒ x = 7

∴ Angles are (6x – 1)˚ : (10x + 2)˚ : (8x + 2)˚ : (9x – 3)˚ : (5x + 4)˚ and (12x + 6)˚

i.e. (6 × 7 – 1)˚ : (10 × 7 + 2)˚ : (8 × 7 + 2)˚ : (9 × 7 – 3)˚ : (5 × 7 + 4)˚ : (12 × 7 + 6)˚

i.e. 41° ; 72°, 58° ; 60° ; 39° and 90°

Let the interior angles of the pentagon be 4x, 5x, 6x, 7x, 5x.

Their sum = 4x + 5x + 6x + 7x + 5x = 21x

Sum of interior angles of a polygon = (n – 2) × 180˚

= (5 – 2) × 180˚

= 540˚

∴ 27x = 540

⇒ x = 540/27

⇒ x = 20˚

∴ Angles are 4 × 20˚ = 80˚

5 × 20˚ = 100˚

6 × 20˚ = 120˚

7 × 20˚ = 140˚

5 × 20˚ = 100˚

⇒ n – 2 = 900/180

⇒ n – 2 = 5

⇒ n = 5 + 2

⇒ n = 7

(ii) Let no. of sides = n

Sum of angles of polygon = 1620˚

(n – 2) × 180˚ = 1620˚

⇒ n – 2 = 1620/180

⇒ n – 2 = 9

⇒ n = 9 + 2

⇒ n = 11

(iii) Let no. of sides = n

Sum of angles of polygon = 16 right = 16 × 90 = 1440˚

(n – 2) × 180˚ = 1440˚

⇒ n – 2 = 1440/180˚

⇒ n – 2 = 8

⇒ n = 8 + 2

⇒ n = 10

(iv) Let no. of sides = n

Sum of angles of polygon = 32 right angles = 32 × 90 = 2880˚

(n × 2) × 180˚ = 2880

n – 2 = 2880/180

n – 2 = 16

n = 16 + 2

n =18

**4. Is it possible to have a polygon; whose sum of interior angles is :****(i) 870°****(ii) 2340°****(iii) 7 right-angles****(iv) 4500°**

**Solution**(i) Let no. of sides = n

Sum of angles = 870°

(n – 2) × 180° = 870°

⇒ n – 2 = 870/180

⇒ n – 2 = 29/6

⇒ n = 29/6 + 2

⇒ n = 41/6

Which is not a whole number.

Hence it is not possible to have a polygon, the sum of whose interior angles is 870°

(ii) Let no. of sides = n

Sum of angles = 2340°

(n – 2) × 180° = 2340°

⇒ n – 2 = 2340/180

⇒ n – 2 = 13

⇒ n = 13 + 2 = 15

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.

(iii) Let no. of sides = n

Sum of angles = 7 right angles = 7 ×90 = 630°

(n – 2) × 180° = 630°

⇒ n – 2 = 630/180

⇒ n – 2 = 7/2

⇒ n = 7/2 + 2

⇒ n = 11/2

Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.

(iv) Let no. of sides = n

(n – 2)×180° = 4500°

⇒ n – 2 = 4500/180

⇒ n – 2 = 25

⇒ n = 25 + 2

⇒ n = 27

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.

**5.****(i) If all the angles of a hexagon are equal; find the measure of each angle.****(ii) If all the angles of a 14-sided figure are equal; find the measure of each angle.**

**Solution**(i) No. of sides of hexagon, n = 6

Let each angle be = x°

Sum of angles = 6x°

(n – 2)× 180° = Sum of angles

⇒ (6 – 2)× 180° = 6x°

⇒ 4× 180 = 6x

⇒ x = (4 × 180)/6

⇒ x = 120˚

∴ Each angle of hexagon = 120˚

(ii) No. of sides of polygon, n = 14

Let each angle = x˚

∴ Sum of angles = 14x˚

∴ (n – 2) × 180˚ = Sum of angles of polygon

∴ (14 – 2) × 180˚ = 14x

12 × 180˚ = 14x

⇒ x = (12 × 180)/14

⇒ x = 1080/7

⇒ x = (154 2/7)˚

**6. Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with :****(i) 7 sides****(ii) 10 sides****(iii) 250 sides.**

**Solution**(i) No. of sides n = 7

Sum of interior & exterior angles at one vertex = 180°

Sum of interior & exterior angles = 7 × 180˚

= 1260˚

Sum of interior angles = (n – 2) × 180˚

= (7 – 2) × 180˚

= 900˚

∴ Sum of exterior angles = 1260˚ - 900˚

= 360˚

(ii) No. of sides n = 10

Sum of interior and exterior angles = 10 × 180˚

= 1800˚

But sum of interior angles = (n – 2) × 180˚

= (10 – 2) × 180˚

= 1440˚

∴ Sum of exterior angles = 1800 – 1440

= 360˚

(iii) No. of sides n = 250

Sum of all interior and exterior angles = 250 × 180˚

= 45000˚

But sum of interior angles = (n – 2) × 180˚

= (250 – 2) × 180˚

= 248 × 180˚

= 44640˚

∴ Sum of exterior angles = 45000 – 44640

= 360˚

**7. The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are (6x – 1)°, (10x + 2)°, (8x + 2)° (9x – 3)°, (5x + 4)° and (12x + 6)° ; find each exterior angle.**

**Solution**Sum of exterior angles of hexagon formed by producing sides of order = 360°

∴ (6x – 1)˚ - (10x + 2)˚ + (8x + 2)˚ + (9x – 3)˚ + (5x + 4)˚ + (12x + 6)˚ = 360˚

50x + 10˚ = 360˚

⇒ 50x = 360˚ - 10˚

⇒ 50x = 350˚

⇒ x = 350/70

⇒ x = 7

∴ Angles are (6x – 1)˚ : (10x + 2)˚ : (8x + 2)˚ : (9x – 3)˚ : (5x + 4)˚ and (12x + 6)˚

i.e. (6 × 7 – 1)˚ : (10 × 7 + 2)˚ : (8 × 7 + 2)˚ : (9 × 7 – 3)˚ : (5 × 7 + 4)˚ : (12 × 7 + 6)˚

i.e. 41° ; 72°, 58° ; 60° ; 39° and 90°

**8. The interior angles of a pentagon are in the ratio 4: 5: 6: 7: 5. Find each angle of the pentagon.**

**Solution**Let the interior angles of the pentagon be 4x, 5x, 6x, 7x, 5x.

Their sum = 4x + 5x + 6x + 7x + 5x = 21x

Sum of interior angles of a polygon = (n – 2) × 180˚

= (5 – 2) × 180˚

= 540˚

∴ 27x = 540

⇒ x = 540/27

⇒ x = 20˚

∴ Angles are 4 × 20˚ = 80˚

5 × 20˚ = 100˚

6 × 20˚ = 120˚

7 × 20˚ = 140˚

5 × 20˚ = 100˚

**9. Two angles of a hexagon are 120° and 160°. If the remaining four angles are equal, find each equal angle.**

**Solution**

Two angles of a hexagon are 120°, 160°

Let remaining four angles be x, x, x and x.

Their sum = 4x + 280°

But sum of all the interior angles of a hexagon

= (6 – 2) × 180˚

= 4 × 180˚

= 720˚

∴ 4x + 280˚ = 720˚

⇒ 4x = 720˚ - 280˚

⇒ 4x = 440˚

⇒ x = 110˚

∴ Equal angles are 110˚

(i) Sum of interior angles of the pentagon

= (5 – 2) × 180˚

= 3 × 180˚ = 540˚

(ii) Since AB ∥ ED

∴ ∠A + ∠E = 180˚

(iii) Let ∠B = 5x ∠C = 6x ∠D = 7x

∴ 5x + 6x + 7x + 180˚ = 540˚

⇒ 18x = 540˚ - 180˚

⇒ 18x = 360˚

⇒ x = 20˚

⇒ x = 110˚

∴ Equal angles are 110˚

**(each)****10. The figure, given below, shows a pentagon ABCDE with sides AB and ED parallel to each other, and ∠B: ∠C : ∠D = 5: 6: 7.****(i) Using formula, find the sum of interior angles of the pentagon.**

**(ii) Write the value of ∠A + ∠E****(iii) Find angles B, C and D.**

**Solution**(i) Sum of interior angles of the pentagon

= (5 – 2) × 180˚

= 3 × 180˚ = 540˚

**[∵ sum for a polygon of x sides = (x – 2) × 180˚]**(ii) Since AB ∥ ED

∴ ∠A + ∠E = 180˚

(iii) Let ∠B = 5x ∠C = 6x ∠D = 7x

∴ 5x + 6x + 7x + 180˚ = 540˚

**(∠A + ∠E = 180˚) Proved in (ii)**⇒ 18x = 540˚ - 180˚

⇒ 18x = 360˚

⇒ x = 20˚

∴ ∠B = 5 × 20˚ = 100˚,

∠C = 6 × 20˚ = 120˚

∠D = 7 × 20 = 140˚

Let number of sides = n

Sum of interior angles = (n – 2) × 180˚

= 180n - 360˚

Sum of 2 right angles = 2 × 90˚

= 180˚

∴ Sum of other angles = 180˚n - 360˚ - 180˚

= 180˚n - 540˚

No. of vertices at which these angles are formed = n – 2

∴ Each interior angle = (180n – 540)/(n – 2)

∴ (180n – 540)/(n – 2) = 120˚

⇒ 180n – 540 = 120n – 240

⇒ 180n – 120n = -240 + 540

⇒ 60n = 300

⇒ n = 300/60

⇒ n = 5

To find: ∠B and ∠D

Proof: No. of sides n = 6

∴ Sum of interior angles = (n – 2) × 180˚

= (6 – 2) × 180˚

= 720˚

∵ AB ∥ EF

∴ ∠A + ∠F = 180˚

But ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 720˚

∠B + ∠C + ∠D + ∠E + ∠180˚ = 720˚

∴ ∠B + ∠C + ∠D + ∠E = 720˚ - 180˚ = 540˚

Ratio = 6 : 4 : 2 : 3

Sum of parts = 6 + 4 + 2 + 3 = 15

∴ ∠B = 6/15 × 540 = 216˚

∠D = 2/15 × 540 = 72˚

Hence, ∠B = 216˚; ∠D = 72˚

Angles of a hexagon are x + 10˚, 2x + 20˚, 2x - 20˚, 3x - 50˚, x + 40˚ and x + 40˚ and x + 20˚

∴ But sum of angles of a hexagon = (x – 2) × 180˚

= (6 – 2) × 180˚

= 4 × 180˚

= 720˚

But sum = x + 10 + 2x + 20˚ + 2x - 20˚ + 3x - 50˚ + x + 40 + x + 20

= 10x + 90 – 70

= 10x + 20

∴ 10x + 20 = 720˚

⇒ 10x = 72 - 20 = 700

⇒ x = 700˚/10 = 70˚

∴ x = 70˚

In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3× 180°

= 540° – 40° – 60° = 540° – 100° = 440°

Ratio in these 3 angles =1 : 3 : 7

Sum of ratios =1 + 3 + 7 = 11

Biggest angle = (440 × 7)/11 = 280°

∠D = 7 × 20 = 140˚

**11. Two angles of a polygon are right angles and the remaining are 120° each. Find the number of sides in it.**

**Solution**Let number of sides = n

Sum of interior angles = (n – 2) × 180˚

= 180n - 360˚

Sum of 2 right angles = 2 × 90˚

= 180˚

∴ Sum of other angles = 180˚n - 360˚ - 180˚

= 180˚n - 540˚

No. of vertices at which these angles are formed = n – 2

∴ Each interior angle = (180n – 540)/(n – 2)

∴ (180n – 540)/(n – 2) = 120˚

⇒ 180n – 540 = 120n – 240

⇒ 180n – 120n = -240 + 540

⇒ 60n = 300

⇒ n = 300/60

⇒ n = 5

**12. In a hexagon ABCDEF, side AB is parallel to side FE and ∠B: ∠C: ∠D: ∠E = 6 : 4: 2: 3. Find ∠B and ∠D.**

**Solution****Given:**Hexagon ABCDEF in which AB∥ EF and ∠B: ∠C: ∠D: ∠E = 6: 4: 2 : 3.To find: ∠B and ∠D

Proof: No. of sides n = 6

∴ Sum of interior angles = (n – 2) × 180˚

= (6 – 2) × 180˚

= 720˚

∵ AB ∥ EF

**(Given)**∴ ∠A + ∠F = 180˚

But ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 720˚

**(Proved)**∠B + ∠C + ∠D + ∠E + ∠180˚ = 720˚

∴ ∠B + ∠C + ∠D + ∠E = 720˚ - 180˚ = 540˚

Ratio = 6 : 4 : 2 : 3

Sum of parts = 6 + 4 + 2 + 3 = 15

∴ ∠B = 6/15 × 540 = 216˚

∠D = 2/15 × 540 = 72˚

Hence, ∠B = 216˚; ∠D = 72˚

**13. The angles of a hexagon are x + 10°, 2x + 20°, 2x – 20°, 3x – 50°, x + 40° and x + 20°. Find x.**

**Solution**Angles of a hexagon are x + 10˚, 2x + 20˚, 2x - 20˚, 3x - 50˚, x + 40˚ and x + 40˚ and x + 20˚

∴ But sum of angles of a hexagon = (x – 2) × 180˚

= (6 – 2) × 180˚

= 4 × 180˚

= 720˚

But sum = x + 10 + 2x + 20˚ + 2x - 20˚ + 3x - 50˚ + x + 40 + x + 20

= 10x + 90 – 70

= 10x + 20

∴ 10x + 20 = 720˚

⇒ 10x = 72 - 20 = 700

⇒ x = 700˚/10 = 70˚

∴ x = 70˚

**14. In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3: 7. Find the biggest angle of the pentagon.**

**Solution**In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3× 180°

= 540° – 40° – 60° = 540° – 100° = 440°

Ratio in these 3 angles =1 : 3 : 7

Sum of ratios =1 + 3 + 7 = 11

Biggest angle = (440 × 7)/11 = 280°

### Exercise 16 B

**1.Fill in the blanks :**

**In case of regular polygon, with :**

**Solution**

**Explanation**

(i) Each exterior angle = 360˚/8 = 45˚

Each interior angle = 180˚ - 45˚ = 135˚

(ii) Each exterior angle = 360˚/12 = 30˚

Each interior angle = 180˚ - 30˚ = 150˚

(iii) Since each exterior = 72˚

∴ Number of sides = 360˚/72˚ = 5

Also interior angle = 180˚ - 72˚ = 108˚

(iv) Since each exterior angle = 45˚

∴ Number of sides = 360˚/45˚ = 8

Interior angle = 180˚ - 45˚ = 135˚

(v) Since interior angle = 150˚

∴ Exterior angle = 180˚ - 150˚ = 30˚

∴ Number of sides = 360˚/30˚ = 12

(vi) Since interior angle = 140˚

∴ Exterior angle = 180˚ - 140˚ = 40˚

∴ Number of sides = 360˚/40˚ = 9

**2. Find the number of sides in a regular polygon, if its each interior angle is :**

(i) 160°

(ii) 135°

(iii) 1 1/5 of a right-angle

Solution

(i) 160°

(ii) 135°

(iii) 1 1/5 of a right-angle

Solution

(i) Let no. of sides of regular polygon be n.

Each interior angle = 160˚

∴ (n – 2)/n × 180˚ = 360˚

⇒ 180˚n - 360˚ = 160n

⇒ 20n = 360˚

⇒ n = 18

(ii) No. of sides = n

Each interior angle = 135˚

(n – 2)/n × 180˚ = 135˚

⇒ 180n - 360˚ = 135n

⇒ 180n – 135n = 360˚

⇒ 45n = 360˚

⇒ n = 8

(iii) No. of sides = n

Each interior angle = 1 1/5 right angles

= 6/5 × 90˚

= 108˚

∴ (n – 2)/n × 180˚ = 108˚

⇒ 180n – 360 ˚ = 108n

⇒ 180n -108n = 360˚

⇒ 72n = 360˚

⇒ n = 5

**3. Find the number of sides in a regular polygon, if its each exterior angle is :**

**(i) 1/3 of a right angle**

**(ii) two-fifth of a right-angle.**

**Solution**

(i) Each exterior angle = 1/3 of a right angle

= 1/3 ×90

= 30°

Let number of sides = n

∴ 360˚/n = 30˚

∴ n = 360˚/30˚

⇒ n = 12

(ii) Each exterior angle = 2/5 of a right-angle

= 2/5 × 90˚

= 36˚

Let number of sides = n

∴ 360˚/n = 36˚

⇒ n = 360˚/36˚

⇒ n = 10

(i) No. of sides = n

each interior angle = 170°

∴ (n – 2)/n × 180˚ = 170˚

⇒ 180n - 360˚ = 170n

⇒ 180n – 170n = 360˚

⇒ 10n = 360˚

⇒ n = 360˚/10˚

⇒ n = 36

which is a whole number.

Hence it is possible to have a regular polygon whose interior angle is 170˚.

(ii) Let no. of sides = n

Each interior angle = 138˚

∴ (n – 2)/n × 180˚ = 138˚

⇒ 180n - 360˚ = 138n

⇒ 180n – 138n = 360˚

⇒ 42n = 360˚

⇒ n = 360˚/42

⇒ n = 60˚/7

Which is not a whole number.

Hence it is not possible to have a regular polygon having interior angle of 138°.

(i) Let no. of sides = n each exterior angle = 80°

360˚/n = 80˚

⇒ n = 360˚/80˚

⇒ n = 9/2

Which is not a whole number.

Hence it is not possible to have a regular polygon whose each exterior angle is of 80°.

(ii) Let number of sides = n

Each exterior angle = 40% of a right angle

= 40/100 × 90

= 36˚

n = 360˚/36˚

⇒ n = 10

Which is a whole number.

Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.

Let each exterior angle or interior angle be = x°

∴ x + x = 180˚

⇒ 2x = 180˚

⇒ x = 90˚

Now, let no. of sides = n

∵ Each exterior angle = 360˚/n

∴ 90˚ = 360˚/n

⇒ n = 360˚/90˚

⇒ n = 4

Let interior angle = x°

Exterior angle = 1/3 x°

∴ x + 1/3x = 180˚

⇒ 3x + x = 540

⇒ 4x = 540

⇒ x = 540/4

⇒ x = 135˚

∴ Exterior angle = 1/3 × 135˚ = 45˚

Let no. of sides = n

∵ Each exterior angle = 360˚/n

∴ 45˚ = 360˚/n

∴ n = 360˚/45˚

⇒ n = 8

Let exterior angle = x°

Interior angle = 5x°

x + 5x = 180°

⇒ 6x = 180°

⇒ x = 30°

Each exterior angle = 30°

Each interior angle = 5 x 30° = 150°

Let no. of sides = n

∵ each exterior angle = 360˚/n

30˚ = 360˚/n

⇒ n = 360˚/30˚

⇒ n = 12

Hence, (i) 150˚ (ii) 30˚ (iii) 12

Interior angle : exterior angle = 2 : 1

Let interior angle = 2x° & exterior angle = x°

∴ 2x˚ + x˚ = 180˚

⇒ 3x = 180˚

Let number of sides = n

∴ 360˚/n = 36˚

⇒ n = 360˚/36˚

⇒ n = 10

**4. Is it possible to have a regular polygon whose each interior angle is :****(i) 170°****(ii) 138°**

**Solution**(i) No. of sides = n

each interior angle = 170°

∴ (n – 2)/n × 180˚ = 170˚

⇒ 180n - 360˚ = 170n

⇒ 180n – 170n = 360˚

⇒ 10n = 360˚

⇒ n = 360˚/10˚

⇒ n = 36

which is a whole number.

Hence it is possible to have a regular polygon whose interior angle is 170˚.

(ii) Let no. of sides = n

Each interior angle = 138˚

∴ (n – 2)/n × 180˚ = 138˚

⇒ 180n - 360˚ = 138n

⇒ 180n – 138n = 360˚

⇒ 42n = 360˚

⇒ n = 360˚/42

⇒ n = 60˚/7

Which is not a whole number.

Hence it is not possible to have a regular polygon having interior angle of 138°.

**5. Is it possible to have a regular polygon whose each exterior angle is :****(i) 80°****(ii) 40% of a right angle.**

**Solution**(i) Let no. of sides = n each exterior angle = 80°

360˚/n = 80˚

⇒ n = 360˚/80˚

⇒ n = 9/2

Which is not a whole number.

Hence it is not possible to have a regular polygon whose each exterior angle is of 80°.

(ii) Let number of sides = n

Each exterior angle = 40% of a right angle

= 40/100 × 90

= 36˚

n = 360˚/36˚

⇒ n = 10

Which is a whole number.

Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.

**6. Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.**

**Solution**Let each exterior angle or interior angle be = x°

∴ x + x = 180˚

⇒ 2x = 180˚

⇒ x = 90˚

Now, let no. of sides = n

∵ Each exterior angle = 360˚/n

∴ 90˚ = 360˚/n

⇒ n = 360˚/90˚

⇒ n = 4

**7. The exterior angle of a regular polygon is one-third of its interior angle. Find the number of sides in the polygon.**

**Solution**Let interior angle = x°

Exterior angle = 1/3 x°

∴ x + 1/3x = 180˚

⇒ 3x + x = 540

⇒ 4x = 540

⇒ x = 540/4

⇒ x = 135˚

∴ Exterior angle = 1/3 × 135˚ = 45˚

Let no. of sides = n

∵ Each exterior angle = 360˚/n

∴ 45˚ = 360˚/n

∴ n = 360˚/45˚

⇒ n = 8

**8. The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find :****(i) measure of each interior angle ;****(ii) measure of each exterior angle and****(iii) number of sides in the polygon.**

**Solution**Let exterior angle = x°

Interior angle = 5x°

x + 5x = 180°

⇒ 6x = 180°

⇒ x = 30°

Each exterior angle = 30°

Each interior angle = 5 x 30° = 150°

Let no. of sides = n

∵ each exterior angle = 360˚/n

30˚ = 360˚/n

⇒ n = 360˚/30˚

⇒ n = 12

Hence, (i) 150˚ (ii) 30˚ (iii) 12

**9. The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :****(i) each exterior angle of the polygon ;****(ii) number of sides in the polygon**

**Solution**Interior angle : exterior angle = 2 : 1

Let interior angle = 2x° & exterior angle = x°

∴ 2x˚ + x˚ = 180˚

⇒ 3x = 180˚

(i) x = 60˚

∴ Each exterior angle = 60˚

Let angle of sides = n

∴ 360˚/n = 60˚

⇒ n = 360˚/60˚

⇒ n = 360˚/60˚

(ii) n = 6

∴ (i) 60˚ (ii) 6

Let exterior angle be x° & interior angle be 4x°

∴ 4x + x = 180˚

⇒ 5x = 180˚

⇒ x = 36˚

∴ Each exterior angle = 36˚

Let no. sides = n

∴ 360˚/n = 36˚

⇒ n = 360˚/36

⇒ n = 10

Let number of sides = n

Sum of exterior angles = 360°

Sum of interior angles = 360° × 2 = 720°

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

720° = (n – 2)× 180°

⇒ n – 2 = 720/180

⇒ n – 2 = 4

⇒ n = 4 + 2

⇒ n = 6

∵ Polygon is regular

∴ AB = BC

⇒ ∠BAC = ∠BAC

But ∠BAC = 20˚

∴ ∠BCA = 20˚

i.e., In Î”ABC,

∠B + ∠BAC + ∠BCA = 180˚

⇒ ∠B + 20˚ + 20˚ = 180˚

⇒ ∠B = 180˚ - 40˚

⇒ ∠B = 140˚

(i) each interior angle = 140˚

(ii) each exterior angle = 180˚ - 140˚ = 40˚

(iii) Let no. of sides = n

∴ 360˚/n = 40˚

⇒ n = 360˚/40˚ = 9

⇒ n = 9

∴ (i) 140˚ (ii) 9

∴ Each exterior angle = 60˚

Let angle of sides = n

∴ 360˚/n = 60˚

⇒ n = 360˚/60˚

⇒ n = 360˚/60˚

(ii) n = 6

∴ (i) 60˚ (ii) 6

**10. The ratio between the exterior angle and the interior angle of a regular polygon is 1: 4. Find the number of sides in the polygon.**

**Solution**Let exterior angle be x° & interior angle be 4x°

∴ 4x + x = 180˚

⇒ 5x = 180˚

⇒ x = 36˚

∴ Each exterior angle = 36˚

Let no. sides = n

∴ 360˚/n = 36˚

⇒ n = 360˚/36

⇒ n = 10

**11. The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.**

**Solution**Let number of sides = n

Sum of exterior angles = 360°

Sum of interior angles = 360° × 2 = 720°

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

720° = (n – 2)× 180°

⇒ n – 2 = 720/180

⇒ n – 2 = 4

⇒ n = 4 + 2

⇒ n = 6

**12. AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find :****(i) its each interior angle,****(ii) its each exterior angle****(iii) the number of sides in the polygon.**

**Solution**∵ Polygon is regular

**(Given)**∴ AB = BC

⇒ ∠BAC = ∠BAC

**[∠s opp. to equal sides]**But ∠BAC = 20˚

∴ ∠BCA = 20˚

i.e., In Î”ABC,

∠B + ∠BAC + ∠BCA = 180˚

⇒ ∠B + 20˚ + 20˚ = 180˚

⇒ ∠B = 180˚ - 40˚

⇒ ∠B = 140˚

(i) each interior angle = 140˚

(ii) each exterior angle = 180˚ - 140˚ = 40˚

(iii) Let no. of sides = n

∴ 360˚/n = 40˚

⇒ n = 360˚/40˚ = 9

⇒ n = 9

∴ (i) 140˚ (ii) 9

**13. Two alternate sides of a regular polygon, when produced, meet at the right angle. Calculate the number of sides in the polygon.**

**Solution**

Let number of sides of regular polygon = n

AB & DC when produced meet at P such that

∠P = 90˚

∵ Interior angles are equal.

∴ ∠ABC = ∠BCD

∴ 180˚ - ∠ABC = 180˚ - ∠BCD

∴ ∠PBC = ∠BCP

But ∠P = 90˚

**(Given)**

∴ ∠PBC + ∠BCP = 180˚ - 90˚ = 90˚

∴ ∠PBC = ∠BCP

= ½ × 90˚

= 45˚

∴ Each exterior angle = 45˚

∴ 45˚ = 360˚/n

⇒ n = 360˚/45˚

⇒ n = 8

**14. In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of:**

**(i) ∠BAE**

**(ii) ∠ABE**

**(iii) ∠BED**

**Solution**

(i) Since number of sides in the pentagon = 5

Each exterior angle = 360/5 = 72°

∠BAE = 180° – 72°= 108°

(ii) In Î”ABE, AB = AE

∴ ∠ABE = ∠AEB

But ∠BAE + ∠ABE + ∠AEB = 180˚

∴ 108˚ + 2∠ABE

= 180˚ - 108˚

= 72˚

⇒ ∠ABE = 36˚

(iii) Since ∠AED = 108˚

**[∵ each interior angle = 108˚]**

⇒ ∠AEB = 36˚

⇒ ∠BED = 108˚ - 36˚

= 72˚

**15. The difference between the exterior angles of two regular polygons, having the sides equal to (n – 1) and (n + 1) is 9°. Find the value of n.**

**Solution**

We know that sum of exterior angles of a polynomial is 360°

**(i)**If sides of a regular polygon = n – 1

Then each angle = 360˚/(n – 1)

And if sides are n + 1, then each angle = 360˚/(n +1)

According to the condition,

360˚/(n – 1) - 360˚/(n + 1) = 9

⇒ 360˚[1/(x - 1) - 1/(x + 1)] = 9

⇒ 360˚ [(n + 1 – n + 1)/(n – 1)(n + 1)] = 9

⇒ (2 × 360)/(n

**– 1) = 9**

^{2}⇒ n

**– 1 = (2×360)/9 = 80**

^{2}⇒ n

**– 1 = 80**

^{2}⇒ n

**= 1 – 80 = 0**

^{2}⇒ n

**– 81 = 0**

^{2}⇒ (n)

**– (9)**

^{2}**= 0**

^{2}⇒ (n + 9)(n – 9) = 0

Either n + 9 = 0, then n = -9 which is not possible being negative,

Or, n – 9 = 0, then n = 9

∴ n = 9

∴ No. of sides of a regular polygon = 9

**16. If the difference between the exterior angle of a n sided regular polygon and an (n + 1) sided regular polygon is 12°, find the value of n.**

**Solution**

We know that sum of exterior angles of a polygon = 360°

Each exterior angle of a regular polygon of 360°

sides = 360˚/n

And exterior angle of the regular polygon of (n + 1) sides = 360˚/(n + 1)

∴ 360˚/n - 360˚/(n + 1) = 12

⇒ 360 [1/n – 1/(n + 1)] = 12

⇒ 360[(n + 1 – n)/n(n + 1)] = 12

⇒ (30 × 1)/(n

**+ n) = 12**

^{2}⇒ 12(n

**+ n) = 360˚**

^{2}⇒ n

**+ n = 36˚**

^{2}**(Dividing by 12)**

⇒ n

**+ n – 30 = 0**

^{2}⇒ n

**+ 6n – 5n – 30 = 0**

^{2}**{∵ -30 = 6 ×(-5) = 1= 6 – 5}**

⇒ n(n + 6) – 5(n + 6) = 0

⇒ (n + 6)(n – 5) = 0

Either n + 6 = 0, then n = - 6 which is not possible being negative

Or, n – 5 = 0, then n = 5

Hence, n = 5

**17. The ratio between the number of sides of two regular polygons is 3:4 and the ratio between the sum of their interior angles is 2:3. Find the number of sides in each polygon.**

**Solution**

Ratio of sides of two regular polygons = 3 : 4

Let sides of first polygon = 3n

and sides of second polygon = 4n

Sum of interior angles of first polygon = (2 × 3n – 4) × 90 ˚

= (6n – 4) × 90˚ and sum of interior angle of second polygon = (2 × 4n – 4) × 90˚

= (8n – 4) × 90˚

∴ ((6n – 4) × 90˚)/((8n – 4) × 90˚) = 2/3

⇒ (6n – 4)/(8n – 4) = 2/3

⇒ 18n – 12 = 16n – 8

⇒ 18n – 16n = -8 + 12

⇒ 2n = 4

⇒ n = 2

∴ No. of sides of first polygon = 3n = 3 × 2 = 6

And no. of sides of second polygon = 4n

= 4 × 2

= 8

**18. Three of the exterior angles of a hexagon are 40°, 51 ° and 86°. If each of the remaining exterior angles is x°, find the value of x.**

**Solution**

Sum of exterior angles of a hexagon = 4 x 90° = 360°

Three angles are 40°, 51° and 86°

Sum of three angle = 40° + 51° + 86° = 177°

Sum of other three angles = 360° – 177° = 183°

Each angle is x°

3x = 183°

⇒ x = 183/3

Hence x = 61

**19. Calculate the number of sides of a regular polygon, if:**

**(i) its interior angle is five times its exterior angle.**

**(ii) the ratio between its exterior angle and interior angle is 2:7.**

**(iii) its exterior angle exceeds its interior angle by 60°.**

**Solution**

Let number of sides of a regular polygon = n

(i) Let exterior angle = x

Then interior angle = 5x

x + 5x = 180°

⇒ 6x = 180°

⇒ x = 180˚/6 = 30˚

∴ Number of sides(n) = 360˚/30 = 12

(ii) Ratio between exterior angle and interior angle = 2 : 7

Let exterior angle = 2x

Then interior angle = 7x

∴ 2x + 7x = 180˚

⇒ 9x = 180˚

⇒ x = 180˚/9

= 20˚

∴ Ext. angle = 2x = 2 × 20˚ = 40˚

∴ No. of sides = 360˚/40 = 9

(iii) Let interior angle = x

Then exterior angle = x + 60

∴ x + x + 60˚ = 180˚

⇒ 2x = 180˚ - 60˚ = 120˚

⇒ x = 120˚/2 = 60˚

∴ Exterior angle = 60˚ + 60˚ = 120˚

∴ Number of sides = 360˚/120˚ = 3

**20. The sum of interior angles of a regular polygon is thrice the sum of its exterior angles. Find the number of sides in the polygon.**

**Solution**

Sum of interior angles = 3 x Sum of exterior angles

Let exterior angle = x

The interior angle = 3x

x + 3x=180°

⇒ 4x = 180°

⇒ x = 180˚/4

⇒ x = 45°

Number of sides = 360/45 = 8

### Exercise 16 C

**1. Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.**

**Solution**

Let the other angle = x°

According to given,

89° + 113° + x° + x° = 360°

⇒ 2x° = 360° – 202°

⇒ 2x° = 158°

⇒ x° = 158/2

other two angles = 79° each

**2. Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5:7; find the measure of each of them.**

**Solution**

Two angles are 68° and 76°

Let other two angles be 5x and 7x

68° + 76°+ 5x + 7x = 360°

⇒ 12x + 144° = 360°

⇒ 12x = 360° – 144°

⇒ 12x = 216°

⇒ x = 18°

angles are 5x and 7x

i.e. 5×18° and 7×18° i.e. 90° and 126°

**3. Angles of a quadrilateral are (4x)°, 5(x + 2)°, (7x – 20)° and 6(x + 3)°. Find :**

**(i) the value of x.**

**(ii) each angle of the quadrilateral.**

**Solution**

Angles of quadrilateral are,

(4x)˚, 5(x + 2)˚, (7x – 20)˚ and 6(x + 3)˚

∴ 4x + 5(x + 2) + (7x – 20)+ 6(x + 3) = 360˚

⇒ 4x + 5x + 10 + 7x – 20 + 6x + 18 = 360˚

⇒ 22x + 8 = 360˚

⇒ 22x = 360˚ - 8˚

⇒ 22x = 352˚

⇒ x = 16˚

Hence angle are,

(4x)˚ = (4 × 16)˚ = 64˚

5(x + 2)˚ = 5(16 + 2)˚ = 90˚

(7x – 20)˚ = (7 × 16 – 20)˚ = 92˚

6(x + 3)˚ = 6(16 + 3) = 114˚

**4. Use the information given in the following figure to find :**

**(i) x**

**(ii) ∠B and ∠C**

**Solution**

∵ ∠A = 90˚

**(Given)**

∠B = (2x + 4˚)

∠C = (3x - 5˚)

∠D = (8x - 15˚)

∠A + ∠B + ∠C + ∠D = 360˚

90˚ + (2x + 4) + (3x – 5) + (8x – 15) = 360

⇒ 90 + 2x + 4 + 3x – 5 + 8x – 15 = 360

⇒ 74˚ + 13x = 360˚

⇒ 13x = 360˚ - 74˚

⇒ 13x = 286˚

⇒ x = 22˚

∵ ∠B = 2x + 4 = 2×22˚ + 4 = 48˚

∠C = 3x – 5 = 3 × 22˚ - 5 = 61˚

Hence (i)22˚ (ii) ∠B = 48˚, ∠C = 61˚

**5. In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4 : 5**

**(i) Calculate each angle of the quadrilateral.**

**(ii) Assign a special name to quadrilateral ABCD**

**Solution**

∵ ∠A : ∠D = 1 : 2

Let ∠A = x and ∠B = 2x

∵ ∠C : ∠B = 4 : 5

Let ∠C = 4y and ∠B = 5y

∵ AB ∥ DC

∴ ∠A + ∠D = 180˚

x + 2x = 180˚

⇒ 3x = 180˚

⇒ x = 60˚

∴ A = 60˚

∠D = 2x = 2 × 60˚ = 120˚

Again ∠B + ∠C = 180˚

5y + 4y = 180˚

⇒ 9y = 180˚

⇒ y = 20˚

∴ ∠B = 5y = 5 × 20˚ = 100˚

∠C = 4y = 4 × 20˚ = 80˚

Hence ∠A = 60˚ : ∠B = 100˚ : ∠C = 80˚ ∠D = 120˚

(iii) Quadrilateral ABCD is a trapezium because one pair of opposite side is parallel.

**6. From the following figure find ;**

**(i) x**

**(ii) ∠ABC**

**(iii) ∠ACD**

**Solution**

(i) In Quadrilateral ABCD,

x + 4x + 3x + 4x + 48° = 360°

⇒ 12x = 360° – 48°

⇒ 12x = 312

⇒ x = 312/12 = 26˚

(ii) ∠ABC = 4x

4 × 26 = 104˚

(iii) ∠ACD = 180˚ - 4x - 48˚

= 180˚ - 4×26 - 48 ˚

= 180˚ - 104˚ - 48˚

= 180˚ - 152˚

= 28˚

∠C = 64°

∠D = ∠C – 8° = 64° - 8° = 56°

∠A = 5(a + 2)°

∠B = 2(2a + 7)°

Now, ∠A + ∠B + ∠C + ∠D = 360°

5(a + 2)° + 2(2a + 7)° + 64° + 56° = 360°

⇒ 5a + 10 + 4a + 14° + 64° + 56° = 360°

⇒ 9a + 144° = 360°

⇒ 9a = 360° – 144°

⇒ 9a = 216°

⇒ a = 24°

∠A = 5 (a + 2) = 5(24 + 2) = 130°

Stun of angles of quadrilateral = 360°

70° + a + 2a + 15 + 3a + 5 = 360°

⇒ 6a + 90° = 360°

⇒ 6a = 270°

⇒ a = 45°

b = 2a + 15 = 2×45 + 15 = 105°

c = 3a + 5 = 3×45 + 5 = 140°

Hence ∠b and ∠c are 105° and 140°

Let each equal angle be x°

x + x + x + 69° = 360°

3x = 360° - 69

⇒ 3x = 291

⇒ x = 97°

Each, equal angle = 97°

∵ ∠P : ∠Q + ∠R : ∠S = 3 : 4 : 5 : 6

Let ∠P = 3x

∠Q = 4x

∠R = 6x

∠S = 7x

∴ ∠P + ∠Q + ∠R + ∠S = 360˚

⇒ 3x + 4x + 6x + 7x = 360˚

⇒ 20x = 360˚

⇒ x = 18˚

∴ ∠P = 3x = 3 × 18 = 54˚

∠Q = 4x = 4 × 18 = 72˚

∠R = 6x = 6 × 18 = 108˚

∠S = 7x = 7 × 18 = 126˚

∠Q + ∠R = 72˚ + 108˚ = 180˚

Or ∠P + ∠S = 54˚ + 126 ˚ = 180˚

Hence PQ ∥ SR

As ∠P + ∠Q = 72˚ + 54˚ = 126˚

Which is ≠ 180˚

∴ PS and QR are not parallel.

4 × 26 = 104˚

(iii) ∠ACD = 180˚ - 4x - 48˚

= 180˚ - 4×26 - 48 ˚

= 180˚ - 104˚ - 48˚

= 180˚ - 152˚

= 28˚

**7. Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(a + 2)° and ∠B = 2(2a + 7)°.****Calculate ∠A.**

**Solution 7:**∠C = 64°

**(Given)**∠D = ∠C – 8° = 64° - 8° = 56°

∠A = 5(a + 2)°

∠B = 2(2a + 7)°

Now, ∠A + ∠B + ∠C + ∠D = 360°

5(a + 2)° + 2(2a + 7)° + 64° + 56° = 360°

⇒ 5a + 10 + 4a + 14° + 64° + 56° = 360°

⇒ 9a + 144° = 360°

⇒ 9a = 360° – 144°

⇒ 9a = 216°

⇒ a = 24°

∠A = 5 (a + 2) = 5(24 + 2) = 130°

**8. In the given figure : ∠b = 2a + 15 and ∠c = 3a + 5; find the values of b and c.****Solution**Stun of angles of quadrilateral = 360°

70° + a + 2a + 15 + 3a + 5 = 360°

⇒ 6a + 90° = 360°

⇒ 6a = 270°

⇒ a = 45°

b = 2a + 15 = 2×45 + 15 = 105°

c = 3a + 5 = 3×45 + 5 = 140°

Hence ∠b and ∠c are 105° and 140°

**9. Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.**

**Solution**Let each equal angle be x°

x + x + x + 69° = 360°

3x = 360° - 69

⇒ 3x = 291

⇒ x = 97°

Each, equal angle = 97°

**10. In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.****Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other****(i) Is PS also parallel to QR?****(ii) Assign a special name to quadrilateral PQRS.**

**Solution**∵ ∠P : ∠Q + ∠R : ∠S = 3 : 4 : 5 : 6

Let ∠P = 3x

∠Q = 4x

∠R = 6x

∠S = 7x

∴ ∠P + ∠Q + ∠R + ∠S = 360˚

⇒ 3x + 4x + 6x + 7x = 360˚

⇒ 20x = 360˚

⇒ x = 18˚

∴ ∠P = 3x = 3 × 18 = 54˚

∠Q = 4x = 4 × 18 = 72˚

∠R = 6x = 6 × 18 = 108˚

∠S = 7x = 7 × 18 = 126˚

∠Q + ∠R = 72˚ + 108˚ = 180˚

Or ∠P + ∠S = 54˚ + 126 ˚ = 180˚

Hence PQ ∥ SR

As ∠P + ∠Q = 72˚ + 54˚ = 126˚

Which is ≠ 180˚

∴ PS and QR are not parallel.

(ii) PQRS is a trapezium as its one pair of opposite side is parallel.

Take A, B, C, D as the vertices of Quadrilateral and BA is produced to E (say).

Since ∠EAD = 70°

∠DAB = 180° – 70° = 110°

⇒ 3x = 360° – 240° = 120°

⇒ x = 40°

Let ∠A = 4x

∠D = 5x

Since ∠A + ∠D = 180°

4x + 5x = 180°

⇒ 9x = 180°

⇒ x = 20°

∠A = 4 (20) = 80°,

∠D = 5 (20) = 100°

Again ∠B + ∠C = 180°

3x – 15° + 4x + 20° = 180°

⇒ 7x = 180° – 5°

⇒ 7x = 175°

⇒ x = 25°

∠B = 75° – 15° = 60°

and ∠C = 4(25) + 20 = 100° + 20° = 120°

The sum of exterior angles of a quadrilateral

y + 80° + 60° + 90° = 360°

⇒ y + 230° = 360°

⇒ y = 360° – 230° = 130°

At vertex A,

∠y + ∠x = 180°

x = 180° – 130°

⇒ x = 50°

The bisector ∠A of the pentagon meets the side CD at point M.

**11. Use the information given in the following figure to find the value of x.****Solution**Take A, B, C, D as the vertices of Quadrilateral and BA is produced to E (say).

Since ∠EAD = 70°

∠DAB = 180° – 70° = 110°

**[EAB is a straight line and AD stands on it ∠EAD + ∠DAB = 180°]**

110° + 80° + 56° + 3x – 6° = 360°**[sum of interior angles of a quadrilateral = 360°]**

3x = 360° – 110° – 80° – 56° + 6°⇒ 3x = 360° – 240° = 120°

⇒ x = 40°

**12. The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.****Solution**Let ∠A = 4x

∠D = 5x

Since ∠A + ∠D = 180°

**[AB || DC]**4x + 5x = 180°

⇒ 9x = 180°

⇒ x = 20°

∠A = 4 (20) = 80°,

∠D = 5 (20) = 100°

Again ∠B + ∠C = 180°

**[ AB || DC]**3x – 15° + 4x + 20° = 180°

⇒ 7x = 180° – 5°

⇒ 7x = 175°

⇒ x = 25°

∠B = 75° – 15° = 60°

and ∠C = 4(25) + 20 = 100° + 20° = 120°

**13. Use the following figure to find the value of x****Solution**The sum of exterior angles of a quadrilateral

y + 80° + 60° + 90° = 360°

⇒ y + 230° = 360°

⇒ y = 360° – 230° = 130°

At vertex A,

∠y + ∠x = 180°

**(Linear pair)**x = 180° – 130°

⇒ x = 50°

**14. ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.**

**Solution****Given:**ABCDE is a regular pentagon.The bisector ∠A of the pentagon meets the side CD at point M.

**To prove:**∠AMC = 90°**Proof:**We know that, the measure of each interior angle of a regular pentagon is 108°.

∠BAM = 1/2 x 108° = 54°

Since, we know that the sum of a quadrilateral is 360°

In quadrilateral ABCM, we have

∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°

⇒ 54° + 108° + 108° + ∠AMC = 360°

⇒ ∠AMC = 360° – 270°

⇒ ∠AMC = 90°

∠1 = ∠4 and ∠3 = ∠5

∠BAM = 1/2 x 108° = 54°

Since, we know that the sum of a quadrilateral is 360°

In quadrilateral ABCM, we have

∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°

⇒ 54° + 108° + 108° + ∠AMC = 360°

⇒ ∠AMC = 360° – 270°

⇒ ∠AMC = 90°

**15. In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively. Show that:****∠AOB = 1/2 (∠C + ∠D)**

**Solution****Given:**AO and BO are the bisectors of ∠A and ∠B respectively.∠1 = ∠4 and ∠3 = ∠5

**...(i)****To prove:**∠AOB = 1/2 (∠C + ∠D)**Proof:**In quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360°

1/2 (∠A + ∠B + ∠C + ∠D) = 180°

Now in ∆AOB

∠1 + ∠2 + ∠3 = 180°

Equating equation (ii) and equation (iii), we get

∠1 + ∠2 + ∠3 = ∠A + ∠B + 1/2 (∠C + ∠D)

⇒ ∠1 + ∠2 + ∠3 = ∠1 + ∠3 + 1/2 (∠C + ∠D)

⇒ ∠2 = 1/2 (∠C + ∠D)

⇒ ∠AOB = 1/2 (∠C + ∠D)

Hence proved.

∠A + ∠B + ∠C + ∠D = 360°

1/2 (∠A + ∠B + ∠C + ∠D) = 180°

**…(ii)**Now in ∆AOB

∠1 + ∠2 + ∠3 = 180°

**…(iii)**Equating equation (ii) and equation (iii), we get

∠1 + ∠2 + ∠3 = ∠A + ∠B + 1/2 (∠C + ∠D)

⇒ ∠1 + ∠2 + ∠3 = ∠1 + ∠3 + 1/2 (∠C + ∠D)

⇒ ∠2 = 1/2 (∠C + ∠D)

⇒ ∠AOB = 1/2 (∠C + ∠D)

Hence proved.