SSC CGL 18 Aug 2021 Shift 2 Paper

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Question : 57 of 100

Marks: +1, -0

Chord

A B

of a circle of radius

10 {cm}

is at a distance

8 {cm}

from the centre

O

. If tangents drawn at

A

and

B

intersect at

P

, then the length of the tangent

A P

(in

{cm}

) is

$7.5 {cm}$
$4 {cm}$
$3.75 {cm}$
$15 {cm}$

Solution:

Given chord

A B=

radius

=10 {cm}

I

△ A O M

,

\;(A O)^{2}=(A M)^{2}+(M O)^{2}

⇒ \;\; (10)^{2}-(8)^{2}=(A M)^{2}

⇒ \;\; A M=√{36}=6 {cm}

\;\text " Now, "\; \tan ∠ A O M=\;{\;\text " Perpendicular "\;}/{\;\text " Base "\;}

=\;\;{A M}/{O M}=\;{6}/{8}=\;{3}/{4}

. . . (i)
Again,

\tan ∠ A O M=\;{A P}/{A O}=\;{A P}/{10}

⇒ \;\; \;{3}/{4}=\;{A P}/{10}

∴ \;\; A P=\;{15}/{2}=7.5 {cm}

[By Eq. (i)]

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