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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

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 1

Consider AB as the diameter with centre O which bisects the chord CD at E

It is given that CE = ED = 8cm and EB = 4cm

Join the diagonal OC

Consider OC = OB = r cm

It can be written as OE = (r – 4) cm

Consider △ OEC

By using the Pythagoras theorem

OC2 = OE2 + EC2

By substituting the values we get

r2 = (r – 4)+ 82

On further calculation r2 = r2 – 8r + 16 + 64

So we get r2 = r2 – 8r + 80

It can be written as r2 – r2 + 8r = 80

We get 8r = 80 By division r = 10 cm

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

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 2

Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.

OA = OB = 5 cm

O'A = O'B = 3 cm

OO' will be the perpendicular bisector of chord AB.

∴ AC = CB

⇒ 52 = AC2 + x2

⇒ 25 − x2 = AC2 ...(1)

In ΔO'AC,

O'A2 = AC2 + O'C2

⇒ 32 = AC2 + (4 − x)2

⇒ 9 = AC2 + 16 + x2 − 8x

⇒ AC2 = − x2 − 7 + 8x ... (2)

From equations (1) and (2), we obtain

25 − x2 = − x2 − 7 + 8x

8x = 32

x = 4

Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.

AC2 = 25 − x2 = 25 − 42 = 25 − 16 = 9

∴ AC = 3 m

Length of the common chord AB = 2 AC = (2 × 3) m = 6 m

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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:

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 3

Using Pythagoras theorem:

OA2 = OM2 + AM2

AM2 = OA2 - OM2

AM2 = 132 - 122

AM2 = 169 - 144

AM2 = 25

AM = 5 cm

AB = 2 AM

AB = 2 x 5 cm = 10 cm

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

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 5

Consider a circle having center O with a chord. Let OA be the radius of the circle and AB be the chord. As given in the question, the radius of the circle is 5 cm and length of chord is 8 cm. Let the distance between the center of the circle and chord be OP. So, this can be shown diagrammatically as:

It is clear from the diagram that OP is perpendicular to AB. As OP is perpendicular to AB and passes through the center O, it will bisect the chord AB at P. Now the length of AP will be,
AP =1/2 × AB

AP = 1/2 × 8

AP = 4 cm
Since, triangle OPA is a right-angle triangle, we can easily apply the Pythagoras theorem which can be stated as b2+p= h2 where b, p and h are base, perpendicular and hypotenuse of the respective triangle.
In ΔOPA,

AP2+OP2=AO2

OP= AO− AP2

OP2= 52−42

OP = 3 cm
Therefore, the distance of the chord AB from the center is 3 cm.

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 = 130° and angle ECD = 20°, then angle BAC is

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 6

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

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 7

In the right triangle OAP,

OA2 = OP2 +AP2 (By Pythagoras theorem)

OA2 = 42 + 32 (perpendicular from the centre of the circle bisects the chord , AP=3cm)

OA2 = 25

OA = 5

Hence the radius of the circle is 5 cm.

Test: Perpendicular From Centre To Chord - Question 8

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

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 8

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

Detailed Solution for Test: Perpendicular From Centre To Chord - Question 9

Here OPA is right angle triangle

given OP = 16 cm

AO = 34 cm

AP = ?

using Pythagoras theorem :

AO² = OP² + AP²

34² = 16² + AP²

1156 = 256 + AP²

1156 - 256 = AP²

900 = AP² 

√900 = AP

30 = AP

Since AD is a straight line and O is at center of both circles

AD = AP + PD

AD = AP + AP

AD = 2AP

AD = 2 x 30

AD = 60 CM

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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