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Test: Perpendicular From Centre To Chord - Grade 9 MCQ


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10 Questions MCQ Test - Test: Perpendicular From Centre To Chord

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Test: Perpendicular From Centre To Chord - Question 1

A circle with centre O in which diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4 cm, the radius of the circle is

Test: Perpendicular From Centre To Chord - Question 2

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. The length of the common chord is

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Test: Perpendicular From Centre To Chord - Question 3

The length of chord which is at a distance of 12 cm from centre of circle of radius 13 cm is:

Test: Perpendicular From Centre To Chord - Question 4

AD is a diameter of a circle and AB is chord. If AB = 24 cm, AD = 30 cm, the distance of AB from the centre of the circle is :

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 4

Test: Perpendicular From Centre To Chord - Question 5

The perpendicular distance of a chord 8 cm long from the centre of a circle of radius 5 cm is

Test: Perpendicular From Centre To Chord - Question 6

A, B, C and D are four points on a circle. AC and BD intersect at E such that angle BEC = 140° and angle ECD = 25°, then angle BAC is

Test: Perpendicular From Centre To Chord - Question 7

The radius of a circle which has a 6 cm long chord, 4 cm away from the centre of the circle is

Test: Perpendicular From Centre To Chord - Question 8

Two concentric circles are intersected by a line L at A, B, C and D. Then

Test: Perpendicular From Centre To Chord - Question 9

In the figure, AD is a straight line. OP is perpendicular to AD and O is the centre of both the circles. If AO = 34 cm, OB = 20 cm and OP = 16 cm, then the length of AD is

Test: Perpendicular From Centre To Chord - Question 10

Two circles of radii10 cm and 8 cm intersect and the length of the common chord is 12 cm. The distance between their centers is

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 10

Given length of common chord AB =12 cm

Let the radius of the circle with centre O is OA = 10 cm

Radius of circle with centre P is AP = 8 cm

From the figure, OP⊥AB

⇒AC = CB

∴AC = 6 cm   (Since AB=12 cm)

In ΔACP, 

AP= PC2+AC2    [By Pythagoras theorem]

⇒ 8= PC2+62  

⇒ PC= 64–36 = 28

 PC = 2√7​ cm

Consider ΔACO,

AO= OC2+AC  [By Pythagoras theorem]

⇒102 = OC2+62  

⇒OC2 = 100−36 = 64

⇒OC = 8 cm
From the figure, OP = OC+PC = 8+2√7​ cm.
Hence, the distance between the centres is (8+2√7​) cm.

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