HOW TO FIND NUMBER OF SIDES OF POLYGON WITH ONE INTERIOR ANGLE

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Regular polygon has all sides equal in length and all angles equal in size.

The sum of interior angles of n sided polygon is 

s = (n - 2) x 180°

Measure of each angle= (n-2)×180n

Problem 1 :

Each interior angle of a regular polygon is 174⁰. Find the number of sides of polygon.

Solution :

174= (n-2)×180n174n180 = n - 287n90 = n - 22 = n -87n902 =90n - 87n902 =3n9n3 = 2n = 2(3)n = 6

So, the number of sides of polygon is 6. Then the polygon is known as hexagon.

Problem 2 :

The interior angle of a regular polygon is 135⁰. Work out the number of sides of the polygon.

Solution :

= (n-2)×180n135n180 = n - 227n36 = n - 22 = n -3n42 =4n - 3n42 =n4n4 = 2n = 2(4)n =

So, the number of sides of the regular polygon is 8.

Problem 3 :

The sum of the interior angles in a polygon is 7380⁰. Calculate the number of sides the polygon has.

Solution :

Sum of the interior angles of a polygon = (n - 2) x 180

(n - 2) x 180 = 7380

n - 2 = 7380/180

n - 2 = 41

Add 2 on both sides, we get

n = 41 + 2

n = 43

Problem 4 :

Shown below is a regular pentagon ABCDE

interiorangleoofpolyq4

Solution :

Number of sides of given polygon = 5

Sum of interior angles = (n - 2) x 180

= (5 - 2) x 180

= 3 x 180

= 540

Measure of each angle = 540/5

x = 108

In triangle BDC,

DC = BC

∠BDC + ∠DCB + ∠CBD = 180

∠BDC = ∠CBD

y + x + y = 180

2y + x = 180

2y + 108 = 180

2y = 180 - 108

2y = 72

Dividing by 2 on both sides.

y = 72/2

y = 36

Problem 5 :

Shown below is a regular hexagon, with an exterior angle labeled y.

interiorangleoofpolyq5

Work out the size of each exterior angle.

Solution :

Number of sides of the polygon above = 6

Sum of the angles of the polygon = (6 - 2) x 180

= 4(180)

= 720

Measure of each angle = 720/6

= 120

Sum of interior angle + y = 180

120 + y = 180

y = 180 - 120

y = 60

Problem 6 :

Shown is a regular hexagon and a regular octagon.

interiorangleoofpolyq6

Solution :

Number of sides of hexagon = 6

Number of sides of octagon = 8

Sum of interior angles of hexagon = (6-2)×1806 = 4×1806= 120 Sum of interior angles of octagon = (8-2)×1808 = 6×1808= 135

y = 360 - (120 + 135)

y = 360 - 255

y = 105

Problem 7 :

Shown below is a regular hexagon ABCDEF.

interiorangleoofpolyq7

Calculate the angle x.

Solution :

Measure of each angle of hexagon = 120

Since it is regular polygon, CD = BC

∠CDB + ∠CBD + ∠BCD = 180

x + x + 120 = 180

2x + 120 = 180

Subtracting 120 on both sides.

2x = 180 - 120

2x = 60

Dividing by 2, we get

x = 60/2

x = 30

Problem 8 :

The diagram below shows a regular pentagon. 𝐴𝐵𝐶 is a straight line

number-of-sides-of-polygon-q1
  • Find 𝑥 Give a reason for your answer.
  • Find 𝑦 Give a reason for your answer.

Solution :

y is the interior angle of the pentagon.

Number of sides of pentagon = 5

= ((n - 2)/n) 180

= (3/5) 180

= 108

y = 108

Sum of angles of x and y is 180.

x + y = 180

x + 108 = 180

x = 180 - 108

x = 72

Problem 9 :

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 polygon = (n - 2) 180

(n - 2) 180 = 2340

n - 2 = 2340/180

n - 2 = 13

n = 13 + 2

n = 15

So, the number of sides of the regular polygon is 15.

Problem 10 :

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

Solution :

Each interior angle = 150

(n - 2) 180/n = 150

180(n - 2) = 150n

180n - 360 = 150n

180n - 150n = 360

30 n = 360

n = 360/30

n = 12

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

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