PROBLEMS ON INTERIOR ANGLES OF A POLYGON

Problem 1 :

Solve for x.

Solution:

Number of sides of the polygon given above = 5

Sum of interior angles of given polygon = (n - 2) × 180°

= (5 - 2) × 180°

= 3 × 180°

= 540°

119° + 90° + 90° + 90° + x° = 540°

389° + x° = 540°

x° = 540° - 389°

x° = 151°

Problem 2 :

Solve for x.

Solution :

Number of sides of the polygon given above = 5

Sum of interior angles of given polygon = (n - 2) × 180°

= (5 - 2) × 180°

= 3 × 180°

= 540°

x° + x° + x° + x° + x° = 540°

 5x° = 540°

x° = 540°/5

 x° = 108°

Problem 3 :

Solve for x.

Solution :

Number of sides of the polygon given above = 5

Sum of interior angles of given polygon = (n - 2) × 180°

= (5 - 2) × 180°

= 3 × 180°

= 540°

90° + (x + 10)° + x° + x° + x° = 540°

 100° + 4x° = 540°

4x° = 540° - 100°

 4x° = 440°

x° = 440/4

x° = 110°

Problem 4 :

The sum of the angles of a polygon is 1980°. How many angles has the polygon?

Solution :

Sum of interior angles of a polygon = 1980°

(n - 2) × 180° = 1980°

n - 2 = 1980/180

n - 2 = 11

n = 11 + 2

n = 13

Hence, the polygon has 13 sides.

Problem 5 :

Juan claims to have found a polygon which has angles with a sum of 2500°. Comment on Juan’s finding.

Solution :

Sum of interior angles of a polygon = 2500°

(n - 2) × 180° = 2500°

n - 2 = 2500/180

n - 2 = 13.8

n = 13.8 + 2

n = 15.8

Since the number of sides would not be decimal, there is no such polygon.

Problem 6 :

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

Solution :

Since the interior and exterior angles is in the ratio of 2 : 7, the angles will be considered as 2x and 7x.

2x + 7x = 180

9x = 180

x = 180/9

x = 20

2(20) ==> 40

7(20) ==> 140

So, the required interior angle exterior angles are 40 and 140 degree respectively.

Problem 7 :

Four angles of a quadrilateral are in the ratio of 3 : 4 : 5 : 6. Find its angles.

Solution :

Let the angles be 3x, 4x, 5x and 6x.

Sum of interior angles of quadrilateral = 360

3x + 4x + 5x + 6x = 360

18x = 360

x = 360/18

x = 20

3x = 3(20) ==> 60

4x = 4(20) ==> 80

5x = 5(20) ==> 100

6x = 6(20) ==> 120

So, the required angles are 60, 80, 100 and 120.

In each diagram below, two regular polygons are shown. Calculate x.

Problem 8 :

Solution :

Number of sides of each polygon = 5

One interior angle measure of pentagon = [(n - 2)/n] 180

= [(5 - 2)/5] 180

= (3/5) 180

= 108

Each interior angle measure of pentagon is 108. Then,

108 + 108 + x = 360

216 + x = 360

x = 360 - 216

x = 144

So, the exterior angle of the polygon shown is 144 degree.

Problem 9 :

Solution :

Number of sides of the polygon which is at left = 5

Number of sides of the polygon which is at right = 8

One interior angle measure of pentagon = [(n - 2)/n] 180

= [(5 - 2)/5] 180

= (3/5) 180

= 108

One interior angle measure of octagon = [(n - 2)/n] 180

= [(8 - 2)/8] 180

= (6/5) 180

= 135

Then,

108 + 135 + x = 360

243 + x = 360

x = 360 - 243

x = 117

So, the exterior angle of the polygon shown is 117 degree.

Problem 10 :

Solution :

Number of sides of the polygon which is at left = 6

Number of sides of the polygon which is at right = 8

One interior angle measure of pentagon = [(n - 2)/n] 180

= [(6 - 2)/6] 180

= (4/5) 180

= 144

One interior angle measure of octagon = [(n - 2)/2] 180

= [(8 - 2)/8] 180

= (6/5) 180

= 135

Then,

144 + 135 + x = 360

279 + x = 360

x = 360 - 279

x = 81

So, the exterior angle of the polygon shown is 81 degree.

Problem 11 :

Solution :

Number of sides of the polygon which is at left = 5

Number of sides of the polygon which is at right = 6

One interior angle measure of pentagon = [(n - 2)/2] 180

= [(5 - 2)/5] 180

= (3/5) 180

= 108

One interior angle measure of octagon = [(n - 2)/2] 180

= [(6 - 2)/6] 180

= (4/5) 180

= 120

Then,

108 + 120 + x = 360

228 + x = 360

x = 360 - 228

x = 132

So, the exterior angle of the polygon shown is 132 degree.