TS ICET 23 May 2018 Shift 2 Paper

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Question : 136 of 200

Marks: +1, -0

As given in the diagram, in △ABC, ∠ B > 90° and AD is perpendicular to BC Then

b^2

=

$a^2 +c^2 -2ax$
$a^2 +c^2 +2ax$
$a^2 +c^2 +ax$
$a^2 +c^2 -ax$

Solution:

In ΔADC

⇒ A C^{2}=A D^{2}+D C^{2}

b^{2}=A D^{2}+(a+x)^{2}

→(1)
In ΔADB

A B^{2}=A D^{2}+B D^{2}

c^{2}=A D^{2}+x^{2}

→ (2)
(1) – (2)

=b^{2}-c^{2}=a^{2}+x^{2}+2 a x-x^{2}

⇒ b^{2}=a^{2}+c^{2}+2 a x

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