SSC CGL 6 06 2019 Shift 2 Paper

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Question : 59 of 100

Marks: +1, -0

A circle is inscribed in AABC, touching AB at P, BC at Q and AC at R. If AR =5 cm, RC =6 cm and AB =12 cm, then the perimeter of AABC is:

40 cm
32 cm
37 cm
36 cm

Solution:

A R=5=A P

P B=7=B Q

R C=6=Q C

SO,

B C=B Q+Q C

=7+6

B C=13

Perimeter

=12+11+13=36 \ {\cm}

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