SSC CHSL 19 Apr 2021 Shift 3 Paper

Section: Quantitative Aptitude

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Question : 54 of 100

Marks: +1, -0

In a right angled triangle

A B C

, if

∠ A B C=90^{∘}, A B=6 {cm}

,

B C=8 {cm}

and

B D

is perpendicular to

A C

, then

A D : D C

is

$7: 16$
$8: 15$
$9: 14$
$9: 16$

Solution:

Given,

In

△ A B C, A C=√{A B^{2}+B C^{2}}

⇒ A C=√{36+64}=√{100}=10 {cm}

In

△ A B C

and

∆ B D C, ∠ B=∠ D=90^{∘}

\;\; B C=B C

(common)

\;\; ∠ C=∠ C

(common)

∴ \;\; ∆ A B C ∼ ∆ B D C

(By ASA criteria)

⇒ \;\; \;{A C}/{B C}=\;{B C}/{D C}

⇒ \;\; D C=\;{B C}/{A C}=\;{64}/{10}

∴ \;\; D C=6.4 {cm}

and

\;\; A D=10-6.4=3.6 {cm}

Now,

A D: D C=3.6: 6.4=9: 16 .

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