SSC CGL 20 Aug 2021 Shift 1 Paper

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Question : 64 of 100

Marks: +1, -0

In

△ A B C, D

is a point on side

A B

such that

B D=3 {cm}

and

D A=4

{cm}

.

E

is a point on

B C

such that

D E \| A C

. Then, area of a

△ B D E

: Area of trapezium

A C E D

is

$40: 9$
$33: 16$
$16: 33$
$9: 40$

Solution:

Given,

A D=4 {cm}, D B=3 {cm}

D E

parallel to

A C

.

\;{∆ B D E}/{∆ B A C}=(\;{\;\text " Side of "\; ∆ B D E}/{\;\text " Side of "\; △ B A C})^{2}

⇒ \;\; \;{∆ B D E}/{∆ B A C}=\;{3^{2}}/{(4+3)^{2}}=\;{9}/{7^{2}}=\;{9}/{49}

∴ \;{\;\text " Area of "\; △ B D E}/{\;\text " Area of trapezium "\; A C E D}

=\;{9}/{\;\text " Area "\;(△ B A C-∆ B D E)}=\;{9}/{49-9}

=\;{9}/{40}=9: 40

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