Quantitative Aptitude Ques 311

Question: 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

Options:

A) $ \frac{a}{2}\sqrt{2a^{2}-b^{2}}sq,units $

B) $ \frac{b}{2}\sqrt{a^{2}-2b^{2}},sq,units $

C) $ \frac{a}{4}\sqrt{4b^{2}-a^{2}}sq,units $

D) $ \frac{b}{4}\sqrt{4a^{2}-b^{2}}sq,units $

Show Answer

Answer:

Correct Answer: C

Solution:

  • By Heron’s formula, Area of $ \Delta =\sqrt{s,(s-a)(s-b)(s-c)} $ and $ s=\frac{a+2b}{2} $ $ \Delta =\sqrt{\frac{a+2b}{2}( \frac{a+2b}{2}-a )( \frac{a+2b}{2}-b )( \frac{a+2b}{2}-b )} $ $ =\sqrt{( \frac{a+2b}{2} )( \frac{2b-a}{2} )(a/2)(a/2)} $ $ =\frac{a}{2}\sqrt{\frac{4b^{2}-a^{2}}{4}}=\frac{a}{4}\sqrt{4b^{2}-a^{2}} $
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