# APPLYING PROPERTIES OF CHORDS WORKSHEET

To learn the properties of chords in a circle,

 Problem 1 : Find the value of y. Problem 2 : Find the value of x. Problem 3 : Find the value of x. Problem 4 : Find the value of x. Problem 5 : Find the value of x. Problem 6 : Find the value of x. Problem 7 : Find the value of x. Problem 8 : Find the value of x.  ## Solutions

Problem 1 :

Find the value of y. Solution : A perpendicular drawn from the center of the circle, bisects the chord.

AE  =  EB  =  15/2  =  7.5 cm

In triangle OEB,

OB2  =  OE2 + EB2

OB2  =  72 + (7.5)2

OB2  =  49 + 56.25

OB2  =  105.25

OB2  =  √105.25

OB  =  10.25 cm

OB  =  OD  =  radius of the circle  =  10.25 cm

In triangle OFD,

OD2  =  OF2 + FD2

(10.25)2  =  72 + FD2

105.0625  =  49 + FD2

Subtract 49 from each side.

56.0625  =  FD2

56.0265  =  FD2

7.49    FD

CD    2(7.49)

CD  ≈  14.98 cm

So, the value of y is about 14.96 cm.

Question 2 :

Find the value of x. Solution :

80°  =  (1/2)[Measure of AB + Measure of CD]

80°  =  (1/2)(x + 60°)

80°  =  (x + 60) / 2

Multiply each side by 2.

160  =  x + 60

Subtract 60 from each side.

100  =  x

Question 3 :

Find the value of x. Solution :

Because AE  =  BD,

ACE  =  BCD

x + 63  =  3x + 1

Subtract x from each side.

63  =  2x + 1

Subtract 1 from each side.

62  =  2x

Divide each side by 2.

31  =  x

Question 4 :

Find the value of x. Solution :

Equal chords of a circle subtend equal angles at the center.

In the above diagram,

ACE  =  BCD

Then

AE  =  BD

2x - 5  =  x

Subtract x from each side.

x - 5  =  0

x  =  5

Question 5 :

Find the value of x. If two chords intersect inside a circle, then the measure of each angle formed is one half the sum of the measures of the arcs intercepted by the angle and its vertical angle.

x  =  (1/2) [measure of arc AB + measure of arc DC]

x  =  (1/2) [80 + 40]

x  =  (1/2)(120)

x  =  60

Question 6 :

Find the values of x. Perpendicular from the centre of a circle to a chord bisects the chord.

AC  =  BC  =  4 cm

In triangle OCB,

OB2  =  OC2 + BC2

x2  =  32 + 42

x2  =  9 + 16

x2  =  25

x  =  √25

x  =  5 cm

Question 7 :

Find the values of x. Solution :

EH ⋅ HG  =  JH ⋅ HF

⋅ 10  =  8 ⋅ x

x  =  (⋅ 10) / 8

x  =  40/8

x  =  5

Question 8 :

AB is a diameter of the circle below. If BC = 2 m and AB = 9 m, find the exact length of AC . Solution :

In triangle ABC,

BCA  =  90°

By Pythagorean theorem,

AB2  =  AC2 + BC2

92  =  AC2 + 22

81  =  AC2 + 4

Subtract 4 from each side.

77  =  AC2

77  =  AC2

77  =  AC Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

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