SSC CGL 8 Aug 2017 Shift 1 Paper

Let the base of the triangle = a cm
∴ Each equal side = $\frac{5}{6}$ a cm
According to the question,
$\mathit{a}+\frac{5}{6}\mathit{a}+\frac{5}{6}\mathit{a}=32$
⇒ $\frac{6\mathit{a}+10\mathit{a}}{6}=32$
⇒ $16\mathit{a}=32\times 6$
∴ a = 12 cm
Hence, the area of the isosceles triangle $=\frac{1}{4}\mathit{b}\sqrt{4{\mathit{a}}^2-{\mathit{b}}^2}$
$=\frac{1}{4}\times 12\sqrt{4{\left(\frac{5}{6}\times 12\right)}^2-{\left(12\right)}^2}$
$=3\sqrt{4\times 100-144}$
$=3\sqrt{400-144}$
= $3\sqrt{256}=3\times 16=48{\mathrm{cm}}^2$