techiesam wrote:

In a triangle ABC, Point D is on side AB and point E is on side AC, such that BCDE is a trapezium. If DE: BC=3:5 then Calculate the ratio of the area of ΔADE and that of trapezium BCED.

A.3:4

B.9:16

C.3:5

D.9:25

E.5:9

Hi,

what you get is Triangle ABC and triangle ADE as two similar triangle..where ABC is the larger triangle..

RATIO of areas is SQUARE of sides of two triangles..SO Area of ADE/Area of ABC = \((\frac{3}{5})^2 = \frac{9}{25}\) = 9x/25x...

But area of BCDE = area of ABC - Area of ADE = 25x-9x = 16x..

so Area of ADE/Area of BCDE = \(\frac{9x}{16x}\) = 9/16...

B

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