Note: Sum of all angles of an n-sided polygon is given by =(n-2)×180

Problem: 01 Find the angle measure of x

Problem: 02 What is the sum of all angles of

(i) a hexagon

(ii) an octagon

(iii) a regular 14 sided polygon

Problem: 03 In the following figures, find the value of x.

Solution:

Problem: 01 Find the angle measure of x

Solution:

The given figure is a regular pentagon

∴ Sum of all angle of pentagon =(n-2)×180=(5-2)×180=540º

Let each measured angle of the pentagon is xº

x+x+x+x+x=540º

5x=540º

x=108º

Thus, each angle measure of the pentagon is 108º

Problem: 02

02 What is the sum of all angles of

(i) a hexagon

(ii) an octagon

(iii) a regular 14 sided polygon

Solution:

The sum of all angles of a polygon is given by (n-2)×180º where n number of side of polygon

(i) We have a hexagon, that has 6 sides

So, the sum of all angles of the hexagon is (6-2)×180º=(4)×180º=720º

(ii) An Octagon have 8 number of side

So, the sum of all angles of the octagon is (8-2)×180º=(6)×180º=1080º

(iii) The sum of all angles of a regular 14 sided polygon is given by (14-2)×180º=(12)×180º=2160º

Problem: 03 In the following figures, find the value of x

Solution:

(i) We know, the sum of all angles of a quadrilateral is 360º

Now, we have 90º,90º,120º, and xº as four angles of the given quadrilateral

∴ 90º+90º+120º+xº=360º

300+x=360º

x=360º-300º=60º

Solution :

(ii) We know, the sum of all angles of a parallelogram (quadrilateral) is 360º

∴ x + 60+x +120+60=360

2x+240=360

2x=360-240

2x=120

x=60º

So, ∠A=60º and ∠B=60+x=60+60=120º

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