SSC CHSL 19 Mar 2020 Shift 1 Paper

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Question : 58 of 100

Marks: +1, -0

In a

\Δ\ABC, \DE

is parallel to BC,

\AD=3 {\cm}, \AE=4 {\cm}

and

\AC=10

{\cm}

, then the value of BD (in

{\cm}

) is

$3.5$
$7.5$
$4.5$
$5.5$

Solution:

Given,

\Δ\DAE=\Δ\BAC

(common)
∵

\DE ∥ \BC

∴

\∠\ADE=\∠\ABC

and

\∠\AED=\∠\ACB

∴

\Δ\ADE \∽\Δ\ABC

[By AA similarity]
∴

{3}/{3+x}={4}/{10}

[\∵

By

{\CPCT}]

⇒

30=12+4 x

⇒

4 x=30-12=18

⇒

x={18}/{4}=4.5 {\cm}

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