INSCRIBED POLYGONS IN CIRCLES WORKSHEET

Problem 1 :

Find the value of x in the diagram shown below. 

Solution :

AB is diameter.  So, ∠C  is a right angle and  m∠C = 90°.

2x°  =  90°

2x  =  90

Divide each side by 2. 

2x/2  =  90/2

x  =  45

Problem 2 :

Find the value of y and z in the diagram shown below. 

Solution :

DEFG is inscribed in a circle, so opposite angles are supplementary.

Problem 3 :

In the diagram, polygon ABCD is inscribed in the circle with center P.  Find the measure of each angle.

Solution :

ABCD is inscribed in a circle, so opposite angles are supplementary.

So, we have 

3x + 3y  =  180 -----(1) 

5x + 2y  = 180 -----(2)

To solve the above system of linear equations, we can solve the first equation for y.

(1)-----> 3x + 3y  =  180

3(x + y)  =  180

Divide each side by 3. 

3(x + y) / 3  =  180 / 3

x + y  =  60

Subtract x from each side. 

y  =  60 - x -----(3)

Plug y  =  60 - x in the second equation. 

(2)-----> 5x + 2(60 - x)  =  180

5x + 120 - 2x  =  180

Simplify.  

3x + 120  =  180

Subtract 120 from each side. 

3x  =  60

Divide each side by 3. 

3x / 3  =  60 / 3

x  =  20

Plug x  =  20 in the third equation. 

(3)-----> y  =  60 - 20

y  =  40

We get x  =  20 and y  =  40. 

So, we have 

m∠A  =  2y°  =  2(40°)  =  80°

m∠B  =  3x°  =  3(20°)  =  60°

m∠C  =  5x°  =  5(20°)  =  100°

m∠D  =  3y°  =  3(40°)  =  120°

Problem 4 :

In figure, if ∠DAB = 60° , ∠ABD = 50°, then find m∠ACB

Solution :

In triangle ADB,

∠DAB + ∠ABD + ∠ADB = 180

60 + 50 + ∠ADB = 180

110 + ∠ADB = 180

∠ADB = 180 - 110

∠ADB = 70

∠ACB = 70

Angle measure created by the same arc will be equal.

Problem 5 :

In figure, O is the centre of the circle ∠BCO = 30°. Find x and y.

Solution :

In triangle OBC,

∠OBC + ∠BCO + ∠BOC = 180

 ∠OBC = ∠BCO

 ∠OBD + y = 30

In triangle OEC,

 ∠OEC + ∠ECO + ∠COE = 180

90 + 30 + ∠COE = 180

 ∠COE = 180 - 120

 ∠COE = 60

 ∠COD = ∠EOD - ∠COE

 ∠COD = 90 - 60

 ∠COD = 30

Angle created at the center is ∠COD = 30

y is the angle created at other point on the circumference of the circle.

2y = ∠COD

2y = 30

y = 30/2

y = 15

In triangle ABE,

x + 30 + 90 + x = 180

2x + 120 = 180

2x = 180 - 120

2x = 60

x = 60/2

x = 30

So, the values of x and y are 30 and 15 respectively.

Problem 6 :

Use the Inscribed Right Triangle-Diameter Theorem to set up and solve an equation to find the value of x.

Solution :

∠NML + ∠LMN + ∠MNL = 180

3x - 6 + 90 + 51 = 180

3x - 6 + 141 = 180

3x + 135 = 180

3x = 180 - 135

3x = 45

x = 45/3

x = 15

So, the value of x is 15.

Problem 7 :

Find the values of x and y. Then find the measures of the interior angles of the polygon.

Solution :

Sum of opposite angles inscribed in the cyclic  quadrilateral = 180

21y + 3x = 180

Dividing by 3, we get

7y + x = 60

x + 7y = 60 ------(1)

26y + 2x = 180

Dividing by 2, we get

x + 13y = 90 -----(2)

(1) - (2)

7y - 13y = 60 - 90

-6y = -30

y = 30/6

y = 5

Applying the value of y, we get

x + 7(5) = 60

x + 35 = 60

x = 60 - 35

x = 25

Problem 8 :

Solution :

14x + 9y = 180 ------(1)

4x + 24y = 180

Dividing by 4, we get

x + 6y = 45 ------(2)

Multiplying (1) by 2 and multiplying (2) by 3 and subtracting, we get

28x + 18y - (3x - 18y) = 360 - 135

28x - 3x + 18y - 18y = 225

25x = 225

x = 225/25

x = 9

Applying the value of x in (2), we get

9 + 6y = 45

6y = 45 - 9

6y = 36

y = 36/6

y = 6

Problem 9 :

Solution :

2x = (1/2)6y

2x = 3y

4x + 6y + 6y = 360

4x + 12y = 360

4x + 3y(4) = 360

4x + 2x (4) = 360

4x + 8x = 360

12x = 360

x = 360/12

x = 30

Applying the value of x, we get

2(30) = 3y

3y = 60

y = 60/3

y = 20

So, the value of x and y are 30 and 20 respectively.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.