SSC CGL 10 Jun 2019 Shift 2 Paper

Question : 60 of 100

Marks: +1, -0

ΔABC ~ ΔEDF and AB = 5 cm , BC = 8 cm and AC = 10 cm . If ar(ΔABC) = ar(ΔDEF) = 9 : 4, then DF is equal to:

Solution:

Since

\Δ {\ABC} \∼\Δ {\EDF}

\ {\text "Area of "\Δ\ {\ABC}}/{\text " Area of " \Δ\ {\DEF}}=\ (\ { {\BC}}/{ {\DF}}\ )^{2}

⇒

\DF =16/3\cm

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