SSC CGL Tier II 25 Oct 2015 Paper 1

Question : 48 of 100

Marks: +1, -0

In trapezium ABCD, AB ||CD and AB = 2CD. Its diagonals intersect at O. If the area of ΔAOB = 84 cm

^{2}

, then the area of ΔCOD is equal to

Solution:

(2)

DC || AB
∠ DCA = ∠ CAB
∠ CDB = ∠ DBA
∴ Δ COD ~ Δ AOB
∴

{\Area \of \Δ\COD}/{\Area \of \Δ\AOB}

= { {CD}^{2}}/{ {AB}^{2}}= { {CD}^{2}}/{4 {CD}^{2}}= {1}/{4}

∴ Area of Δ COD

=1/4×84=21

sq. cm.

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