# Frank Solutions for Chapter 18 Rectilinear Figures Class 9 Mathematics ICSE

### Exercise 18.1

1. Find the sum of the interior angles of a polygon of:

(i) 7 sides

(ii) 12 sides

(iii) 9 sides

(i) When n = 7

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

= (7 – 2)×180°

On further calculation, we get,

= 5×180°

= 900°

(ii) When n = 12

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

= (12 – 2)×180°

On further calculation, we get,

= 10×180°

= 1800°

(iii) When n = 9

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

= (9 – 2)×180°

On further calculation, we get,

= 7×180°

= 1260°

2. Find the measure of each interior angle of a regular polygon of:

(i) 6 sides

(ii) 10 sides

(iii) 15 sides

(i) When n = 6

∴ Each interior angle of a regular polygon = {(n – 2)×180°}/n

= {(6 – 2)×180°}/6

On further calculation, we get,

= (4×180°)/6

= 120°

(ii) When n = 10

∴ Each interior angle of a regular polygon = {(n – 2)×180°}/n

= {(10 – 2)×180°}/10

On further calculation, we get,

= (8×180°)/10

= 144°

(iii) When n = 15

∴ Each interior angle of a regular polygon = {(n-2)×180°}/n

= {(15 – 2)×180°}/15

On further calculation, we get,

= (13×180°)/15

= 156°

3. Find each exterior angle of a regular polygon of:

(i) 9 sides

(ii) 15 sides

(iii) 18 sides

(i) When n = 9

∴ Each exterior angle of a regular polygon = (360°/n)

= (360°/9)

We get,

= 40°

(ii) When n = 15

∴ Each exterior angle of a regular polygon = (360°/n)

= (360°/15)

We get,

= 24°

(iii) When n = 18

∴ Each exterior angle of a regular polygon = (360°/n)

= (360°/18)

We get,

= 20°

4. Find the number of sides in a regular polygon, when each interior angle is:

(i) 120°

(ii) 140°

(iii) 135°

(i) Each interior angle of a regular polygon = {(n-2)×180°}/n

{(n - 2)×180°}/n = 120°

On calculating further, we get,

180° (n – 2) = 120° (n)

⇒ 3 (n – 2) = 2n

⇒ 3n – 6 = 2n

⇒ 3n – 2n = 6

We get,

n = 6

(ii) Each interior angle of a regular polygon = {(n-2) x 180°}/n

{(n – 2)×180°}/n = 140°

On calculating further, we get,

180° (n – 2) = 140(n)

⇒ 9 (n – 2) = 7n

⇒ 9n – 18 = 7n

⇒ 9n – 7n = 18

⇒ 2n = 18

We get,

n = 9

(iii) Each interior angle of a regular polygon = {(n – 2)×180°}/n

{(n – 2)× 180°}/n = 135°

On calculating further, we get,

180° (n – 2) = 135° (n)

⇒ 4 (n – 2) = 3n

⇒ 4n – 8 = 3n

⇒ 4n – 3n = 8

We get,

n = 8

5. Find the number of sides in a regular polygon, when each exterior angle is:

(i) 20°

(ii) 60°

(iii) 72°

(i) Each exterior angle = (360°/n)

⇒ 360°/n = 20°

We get,

n = 18

(ii) Each exterior angle = (360°/n)

⇒ 360°/n = 60°

We get,

n = 6

(iii) Each exterior angle = (360°/n)

⇒ 360°/n = 72°

We get,

n = 5

6. The angles of a pentagon are 100°, 96°, 74°, 2x°and 3x°. Find the measures of the two angles 2x°and 3x°.

We know that,

A pentagon has 5 sides

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

= (5 – 2)×180°

On calculating further, we get,

= 3×180°

We get,

= 540°

Given angles are 100°, 96°, 74°, 2x° and 3x°

∴ 100° + 96° + 74° + 2x° + 3x° = 540°

5x° + 270° = 540°

⇒ 5x° = 540° – 270°

⇒ x° = (540° – 270°)/5

We get,

x° = 54°

Hence,

2x° = 2×54 = 108° and 3x° = 3×54 = 162°

Therefore, the two angles are 108° and 162° respectively.

7. The three angles of a quadrilateral are 71°, 110°, 95°. Find its fourth angle.

A quadrilateral is a polygon with four sides

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

= (4 – 2)×180°

= 2×180°

We get,

= 360°

Given, the three interior angles are 71°, 110°, 95°

Let x be the fourth angle

∴ 71° + 110° + 95° + x = 360°

x + 276° = 360°

We get,

x° = 360° – 276°

We get,

= 84°

Hence, the fourth angle is 84°

8. Find the angles of a pentagon which are in the ratio 4: 4: 6: 7: 6.

A pentagon has 5 sides

∴ Sum of interior angles = (n-2)×180°

= (5 – 2)×180°

= 3×180°

We get,

= 540°

Ratio of the angles are 4x°, 4x°, 6x°, 7x° and 6x°

∴ 4x° + 4x° + 6x° + 7x° + 6x° = 540°

⇒ 27x° = 540°

We get,

x° = 20°

Therefore, the interior angles of a pentagon are 80°, 80°, 120°, 140° and 120°

9. Find the angles of a quadrilateral whose angles are in the ratio 1: 4: 5: 2.

A quadrilateral is a polygon with four sides

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

= (4 – 2)×180°

= 2×180°

We get,

= 360°

Given ratio of the angles = 1: 4: 5: 2

Hence,

The interior angles are x°, 4x°, 5x° and 2x°

x° + 4x° + 5x° + 2x° = 360°

⇒ 12x° = 360°

We get,

x° = 30°

Therefore, the interior angles of the quadrilateral are 30°, 120°, 150° and 60°

10. The angles of a pentagon are x°, (x – 10)°, (x + 20°), (2x – 44)°and (2x – 70)°. Find the angles.

A pentagon has 5 sides

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

= (5 – 2)×180°

= 3×180°

We get,

= 540°

Given angles are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70)°

∴ x° + (x – 10)° + (x + 20)° + (2x – 44)° + (2x – 70°) = 540°

⇒ 7x° – 104° = 540°

⇒ 7x° = 540° + 104°

⇒ 7x° = 644°

We get,

x° = 92°

Therefore, the interior angles of the pentagon are 92°, 82°, 112°, 140° and 114°

11. The angles of a hexagon are (2x + 5)°, (3x – 5)°, (x + 40)°, (2x + 20)°, (2x + 25)°and (2x + 35)°. Find the value of x.

A hexagon has 6 sides

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

= (6 – 2)×180°

= 4×180°

We get,

= 720°

Given angles of a hexagon are (2x + 5)°, (3x – 5)°, (x + 40)°, (2x + 20)°, (2x + 25)° and (2x + 35)°

∴ (2x + 5)° + (3x – 5)° + (x + 40)° + (2x + 20)° + (2x + 25)° + (2x + 35)° = 720°

⇒ 12x + 120° = 720°

We get,

x = 50°

12. One angle of a hexagon is 140°and the remaining angles are in the ratio 4: 3: 4: 5: 4. Calculate the measures of the smallest and the largest angles.

A hexagon has 6 sides

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

= (6 – 2)×180°

= 4×180°

We get,

= 720°

One angle of a hexagon is given to be 140°

Ratio of the remaining five angles = 4: 3: 4: 5: 4

∴ The interior angles are 4x°, 3x°, 4x°, 5x° and 4x°

∴ 140° + 4x° + 3x° + 4x° + 5x° + 4x° = 720°

⇒ 20x° + 140° = 720°

⇒ x° = 580°/20

We get,

x° = 29°

Smallest angle is 3x° = 3×29° = 87° and

Largest angle is 5x° = 5×29° = 145°

Therefore, the smallest angle is 87° and the largest angle is 145°

13. One angle of a pentagon is 160°and the rest are all equal angles. Find the measure of the equal angles.

A pentagon has 5 sides

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

= (5 – 2)×180°

= 3×180°

We get,

= 540°

One angle of a pentagon is given to be 160°

Ratio of the remaining four angles = 1: 1: 1: 1

∴ The interior angles are x°, x°, x° and x°

∴ 160° + x° + x° + x° + x° = 540°

⇒ 4x° = 540° – 160°

⇒ 4x° = 380°

We get,

x° = 95°

Therefore, each equal angle is 95°

14. Calculate the measure of each angle of a nonagon.

A nonagon has 9 sides

∴ Each interior angle of a regular polygon = {(n – 2)×180°}/n

= {(9 – 2)×180°}/9

= (7×180°)/9

We get,

= 140°

15. Calculate the measure of each angle of a regular polygon of 20 sides.

Given

n = 20

∴ Each interior angle of the regular polygon = {(n – 2) x 180°}/n

= {(20 – 2)×180°}/20

= (18×180°)/20

We get,

= 162°

16. (a) It is possible to have a polygon whose sum of interior angles is 780° ?

(b) Is it possible to have a polygon whose sum of interior angles is 7 right angles?

(a)

(b)

17. (a) Is it possible to have a polygon whose each interior angle is 124°?

(b) Is it possible to have a polygon whose each interior angle is 105°?

(a)

(b)

18. A heptagon has three angles equal to 120°, and the other four angles are equal. Find all the angles.

19. In a pentagon ABCDE, AB || ED and B = 140°C = 2xand D = 3x°. Find C and D.

20. In a polygon, there are 3 right angles and the remaining angles are equal to 165°. Find the number of sides in the polygon.

21. ABCDE is a pentagon in which AB is parallel to DC and A : E : D = 1 : 2 : 3. Find angle A.

22. If the difference between an exterior angle of a regular polygon of ‘n’ sides and an exterior angles of another regular polygon of (n + 1)’ sides is equal to 4°; find the value of ‘n’.

23. The number of sides of two regular polygon are in the ratio 2 : 3 and their interior angles are in the ratio 9 : 10. Find the number of sides of each polygon.

24. KL, LM and MN are three consecutive sides of a regular polygon. If LKM = 20°, find the interior angle of the polygon and the number of sides of the polygon.

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

26. Find the value of each angle of a heptagon if three of its angles measure 132° each and the remaining four.

27. Find the value of each angle of an octagon if two of its angles are 148° and 152° and the remaining angles are all equal.

28. Find the value of each angle of an octagon if four of its angles are equal and the other four are each greater than these by 20°.

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

30. The interior angle of a regular polygon is double the exterior angle. Find the number of sides of the polygon.

31. The sum of the interior angles of a polygon is 6.5 times the sum of its exterior angles. Find the number of sides of the polygon.

32. The difference between an exterior angle of (n – 1) sides regular polygon and an exterior angle of (n + 2) sided regular polygon is 6°. Find the value of n.

33. In a pentagon PQRST, P = 100°Q = 120and S = T. The sides PQ and SR, when produced meet at right angle. Find QRS and PTS.

34. In a hexagon JKLMNO, side JK || ON and K : L : M : N = 6 : 5 : 4 : 3. Find the angle K and M.

35. In a regular pentagon PQRST, PR = QT intersect at N. Find the angle RQT and QNP.