FIND THE NUMBER OF SIDES OF THE POLYGON IF EXTERIOR ANGLE GIVEN

Problem 1 :

Find the number of sides of a regular polygon whose each exterior angle measures 60°.

Solution :

Sum of all exterior angles = 360°

n × 60 = 360

n = 360/60

n = 6

Hence, the number of sides in the polygon is 6.

Problem 2 :

An exterior angle and the interior angle of a regular polygon are in the ratio 2:7. Find the number of sides of the polygon.

Solution :

Let x be the interior angle, then 180-x be the exterior angle.

(180-x) : x = 2 : 7

(180-x)/x = 2 / 7

2x = 7(180-x)

2x = 1260 - 7x

2x + 7x = 1260

9x = 1260

x = 1260/9

x = 40

Exterior angle is 40 degree. Interior angle = 140

140 = (n - 2)(180/n)

140n = 180(n - 2)

140n = 180n - 360

140n - 180n = -360

-40n = -360

n = 360/40

n = 9

Hence, the number of sides in the polygon is 9.

Problem 3 :

Each exterior angle of a regular polygon is 20⁰. Work out the number of sides of the polygon.

Solution :

Exterior angle of the regular polygon = 20

Each interior angle = 180 - 20

= 160

Measure of each interior angle = [(n - 2) 180]/n

160 = (180/n)(n - 2)

160n/180 = n - 2

8n/9 = n - 2

8n = 9(n - 2)

8n = 9n - 18

8n - 9n = -18

-n = -18

n = 18

So, the number of sides of the polygon is 18.

Problem 4 :

The number of sides of a regular polygon whose each exterior angle is 60° is

Solution :

Exterior angle of the regular polygon = 60

Each interior angle = 180 - 60

= 120

Measure of each interior angle = [(n - 2) 180]/n

120 = (180/n)(n - 2)

120n/180 = n - 2

2n/3 = n - 2

2n = 3(n - 2)

2n = 3n - 6

2n - 3n = -6

-n = -6

n = 6

So, the number of sides of the polygon is 6.

Problem 5 :

The interior angle of a regular polygon is four times its exterior angle. How many sides does the polygon have ?

Solution :

Let x be the interior angle, then 180 - x be the exterior angle.

x = 4(180 - x)

x = 720 - 4x

Adding 4x, we get

x + 4x = 720

5x = 720

x = 720/5

x = 144

Interior angle = 144.

Measure of each interior angle = [(n - 2) 180]/n

144 = (180/n)(n - 2)

1440n/180 = n - 2

4n/5 = n - 2

4n = 5(n - 2)

4n = 5n - 10

4n - 5n = -10

-n = -10

n = 10

So, the number of sides of the polygon is 10.

Problem 6 :

Find the number of sides for a regular polygon whose measure of each interior angle is 150°.

Solution :

One interior angle of polygon = 150

(n - 2) (180/n) = 150

180(n - 2) = 150n

180n - 360 = 150n

180n - 150n = 360

30n = 360

n = 360/30

n = 12

So, the required number of sides of the polygon is 12.

Problem 7 :

Find the number of sides for a regular polygon whose measure of each interior angle is 168°.

Solution :

One interior angle of polygon = 168

(n - 2) (180/n) = 168

180(n - 2) = 168n

180n - 360 = 168n

180n - 168n = 360

12n = 360

n = 360/12

n = 30

So, the required number of sides of the polygon is 30.

Problem 8 :

Find the number of sides for a regular polygon whose sum of the measures of its interior angles is 2340°.

Solution :

Sum of interior angles of the polygon = 2340

(n - 2) 180 = 2340

n - 2 = 2340/180

n - 2 = 13

n = 13 + 2

n = 15

Problem 9 :

A heptagon has seven sides. What is the sum of the measures of its interior angles?

Solution :

Number of sides of heptagon = 7

Sum of the measures of interior angle = (n - 2) 180

= (7 - 2) 180

= 5 (180)

= 900

Problem 10 :

ABCDE is a regular pentagon. DEF is a straight line. Calculate

(a) angle AEF

(b) angle DAE.

Solution :

(a) angle AEF

Number of sides of pentagon = 5

angle AEF = (n - 2) (180/n)

= (5 - 2)(180/5)

= 3(36)

= 108

(b) angle DAE

ED = EA

∠DAE = x = ∠ADE

x + x + 108 = 180

2x + 108 = 180

2x = 180 - 108

2x = 72

x = 72/2

x = 36

angle DAE = 36