SSC CGL 23 Aug 2021 Shift 1 Paper

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Question : 66 of 100

Marks: +1, -0

A circle is incribed in quadrilateral

A B C D

, touching sides

A B, B C, C D

and

D A

at

P, Q, R

and

S

, respectively. If

A S=6 {cm}

,

B C=12 {cm}

and

C R=5 {cm}

, then the length of

A B(

in

{cm})

is

15
11
12
13

Solution:

Given

A S=6 {cm}, B C=12 {cm}

,

C R=5 {cm}

If

A S=6 {cm}

then

A P

is also

6 {cm}

\{ ∵

both are tangents

\}

Similarly,

B C=12 {cm}

and

C R=5 {cm}

C R=C Q=5 {cm} \;\text ". "\;

So,

B Q=12-5=7 {cm}

and

B P=B Q=7 {cm}

So,

A B=A P+P B

=6+7=13 {cm}

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