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:
Option 1: $\frac{1}{4}\sqrt{4b^{2}-a^{2}}$ sq. units
Option 2: $\frac{a}{2}\sqrt{2a^{2}-b^{2}}$ sq. units
Option 3: $\frac{b}{4}\sqrt{4a^{2}-b^{2}}$ sq. units
Option 4: $\frac{b}{2}\sqrt{a^{2}-2b^{2}}$ sq. units
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Correct Answer: $\frac{b}{4}\sqrt{4a^{2}-b^{2}}$ sq. units
Solution :
In an isosceles triangle with two sides of length $a$ units and a base of length $b$ units.
The height can be found using the Pythagorean theorem,
$ = \sqrt{a^2 - \left(\frac{b}{2}\right)^2}$
The area of the triangle,
$= \frac{1}{2} \times$ base $\times$ height
$= \frac{1}{2} \times b \times \sqrt{a^2 - \left(\frac{b}{2}\right)^2}$
$= \frac{1}{2} \times b \times \sqrt{a^2 - \left(\frac{b^2}{4}\right)}$
$= \frac{1}{2} \times b \times \sqrt{\left(\frac{4a^2-b^2}{4}\right)}$
$= \frac{1}{4} \times b \times \sqrt{(4a^2-b^2)}$
$= \frac{b}{4} \times \sqrt{(4a^2-b^2)}$
Hence, the correct answer is $ \frac{b}{4} \times \sqrt{(4a^2-b^2)}$ sq. units.
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