# 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**

**Answer**

**(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**

**Answer**

**(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**

**Answer**

**(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****°**

**Answer**

**(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^{0 }(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°**

**Answer**

**(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°.**

**Answer**

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.**

**Answer**

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,

x°^{ }= 84°

Hence, the fourth angle is 84°

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

**Answer**

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.**

Answer

Answer

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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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.**

**Answer**

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?**

**Answer**

(a)

**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°?**

**Answer**

(a)

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

**Answer**

**19. In a pentagon ABCDE, AB ****|| ED and ****∠****B = 140****°****, ****∠****C = 2x**^{0 }**and ****∠****D = 3x****°****. Find ****∠****C and ****∠****D.**

**Answer**

**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.**

**Answer**

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

**Answer**

**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’.**

**Answer**

**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.**

**Answer**

**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.**

**Answer**

**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.**

**Answer**

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

**Answer**

**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.**

**Answer**

**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°.**

**Answer**

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

**Answer**

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

**Answer**

**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.**

**Answer**

**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.**

**Answer**

**33. In a pentagon PQRST, ****∠****P = 100****°****, ****∠****Q = 120**^{0 }**and ****∠****S = ****∠****T. The sides PQ and SR, when produced meet at right angle. Find ****∠****QRS and ****∠****PTS.**

**Answer**

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

**Answer**

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

**Answer**

**36. Each exterior angle of a regular polygon is 1/P times of its interior angle. Find the number of sides in the polygon.**

**Answer**

**37. Each exterior angle of a regular polygon is 162°. Another regular polygon has number of sides double the first polygon. Find each interior angle of the second polygon.**

**Answer**