APPLYING PROPERTIES OF CHORDS WORKSHEET

About "Applying properties of chords worksheet"

Applying properties of chords worksheet :

Here we are going to see some practice questions on applying properties of chords.

Applying properties of chords worksheet - Practice questions

(1)  Find the value of "y".

(2)  Find the value of "x".

(3)  Find the value of "x".

(4)  Find the value of "x".

(5)  Find the value of "x".

(6)  Find the value of "x".

(7)  Find the value of "x".

(8)  Find the value of "x".

Question 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

OB  =  √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

(10.25)2  -  72  =  FD2

FD2  =  105.06 - 49

FD2  =  56.06

FD  =  √56.06

FD  =  7.48 cm

CD  =  2(7.48)  =  14.96 cm

Hence the value of y is 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 (2)  =  x + 60

160 - 60  =  x

x  =  100°

Hence the value of x is 100°.

Question 3 :

Find the value of x. 

Solution :

<ACE  =  <BCD

x + 63  =  3x + 1

x - 3x  =  1 - 63

-2x  =  -62

x  =  62/2

x  =  31

Hence the value of x is 31.

Question 4 :

Find the value of "x".

Solution :

Property going to be used :

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

Since AE  =  BD,

<ACE  =  <BCD

2x - 5  =  x

2x - x  =  5

x  =  5

Hence the value of x is 5.

Question 5 :

Find the value of "x".

Property going to be used :

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° 

Hence the value of x is 60°.

Question 6 :

Find the values of "x".

Property going to be used :

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 cm

Hence the value of x is 5 cm.

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° 

AB2  =  AC2 + BC2

92  =  AC2 + 22

81 - 4  =  AC2

AC  =  √77

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