SSC CGL 10 Jun 2019 Shift 1 Paper

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Question : 72 of 100

Marks: +1, -0

In ΔADC, E and B are the points on the sides AD and AC respectively such that ∠ABE = ∠ADC. If AE = 6cm, BC=2cm, BE = 3cm and CD = 5 cm, then (AB +DE) is equal is:

14 cm
16 cm
$49/3$ cm
$46/3$ cm

Solution:

\ln \Δ\ {\ABE}

and

\Δ\ {\ADC}

\∠{A}

is common

\∠\ {\ABE}=\∠\ {\ADC}(

Given

)

\ {\So}, \∠\ {\AEB}=\∠\ {\ACD}

\∴\Δs

are similar
Now

\ {A E}/{A C}=\ {E B}/{D C}=\ {A B}/{A D}

\ {6}/{A C}=\ {3}/{5} \⇒A C=10

units
Now

\ {A B}/{A D}=\ {E B}/{D C}

\ {8}/{D E+6}=\ {3}/{5}

40=3 D E+18

\ {DE}=\ {22}/{3}

Now,

A B+D E=8+\ {22}/{3}=\ {46}/{3} \ {\cm}

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