SSC CGL 11 Jun 2019 Shift 1 Paper

Question : 75 of 100

Marks: +1, -0

ΔABC ~ ΔQRP and PQ = 6 cm, QR = 8 cm and PR = 10 cm. If ar(ΔABC) : ar(ΔPQR) = 1:4, then AB is equal to:

Solution:

\ {\since} \ΔA B C \∼\ΔQ R P

\ {\ {\ar}(\ΔA B C)}/{\ {\ar}(\ΔP Q R)}=\ {A B^{2}}/{Q R^{2}}

\⇒\ {1}/{4}=\ {A B^{2}}/{8^{2}}

\⇒A B=4 \ {\cm}

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