| 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 [SSC (CGL) Mains 2014] |
A) \[\frac{a}{2}\sqrt{2{{a}^{2}}-{{b}^{2}}}\,\,\text{sq}\,\text{units}\]
B) \[\frac{b}{2}\sqrt{{{a}^{2}}-2{{b}^{2}}}\,\text{sq}\,\text{units}\]
C) \[\frac{a}{4}\sqrt{4{{b}^{2}}-{{a}^{2}}}\,\,\text{sq}\,\text{units}\]
D) \[\frac{b}{4}\sqrt{4{{a}^{2}}-{{b}^{2}}}\,\,\text{sq}\,\text{units}\]
Correct Answer: C
Solution :
| By Heron's formula, |
| Area of \[\Delta =\sqrt{s\,(s-a)(s-b)(s-c)}\] and \[s=\frac{a+2b}{2}\] |
| \[\Delta =\sqrt{\frac{a+2b}{2}\left( \frac{a+2b}{2}-a \right)\left( \frac{a+2b}{2}-b \right)\left( \frac{a+2b}{2}-b \right)}\]\[=\sqrt{\left( \frac{a+2b}{2} \right)\left( \frac{2b-a}{2} \right)(a/2)(a/2)}\] |
| \[=\frac{a}{2}\sqrt{\frac{4{{b}^{2}}-{{a}^{2}}}{4}}=\frac{a}{4}\sqrt{4{{b}^{2}}-{{a}^{2}}}\] |
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