SSC CGL Tier II 12 Apr 2015 Paper

Question : 11 of 100

Marks: +1, -0

If for an isosceles triangle the length of each equal side is ‘a’ units and that of the third side is ‘b’ units, then its area will be

Solution:

(3)

AD, is perpendicular on BC.
BD = DC =

\b/2

{AD}=√{ {AB}^{2}- {BD}^{2}}

=√{ {a}^{2}- ( { {b}}/{2} )^{2}}= √{ {a}^{2}-\ {b^{2}}/{4}}

= {√{4 a^{2}-b^{2}}}/{2}

units

∴

Area of

\Δ A B C

=\ {1}/{2} ×B C × A D

= {1}/{2} b×{√{4 a^{2}-b^{2}}}/{2}

= {b}/{4} √{4 a^{2}-b^{2}}

square units.

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