SSC CGL Tier II 25 Oct 2015 Paper 1

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Question : 14 of 100

Marks: +1, -0

In ΔABC ∠BAC = 90° and AD ⊥ BC. If BD = 3 cm and CD = 4 cm, then the length of AD is

3.5 cm
5 cm
$2 √3$ cm
6 cm

Solution:

(3)

BD = 3 cm
CD = 4 cm
In ΔABC,

A B^{2}+A C^{2}=7^{2}

⇒ A B^{2}+A C^{2}=49 ....(i)

In ΔABD,

{AB}^{2}= {AD}^{2}+3^{2}= {AD}^{2}+9 ...(ii)

In ΔADC,

{AC}^{2}= {AD}^{2}+16....(iii)

On adding equations

(ii)

and

(iii)

,

A B^{2}+A C^{2}=A D^{2}+9+A D^{2}+16

⇒ 49=2 A D^{2}+25

⇒ 2 A D^{2}=49-25=24

⇒ A D^{2}=12

⇒ A D=√{12}=2√{3} {cm}

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