SSC CPO SI and ASI 2012 Paper

(1) Let the two sides of triangle be b
and c, respectively.Given, Perimeter of triangle = 32 cm.
So, 32 = 11 + b + c
⇒ b + c = 21
Also, b – c = 5. Therefore b = 13 cmand c = 8 cm
Now, Area of triangle
$=\sqrt{\mathrm{}\mathit{s}\left(\mathrm{}\mathit{s}-\mathit{a}\right)\left(\mathrm{}\mathit{s}-\mathit{b}\right)\left(\mathit{s}-\mathit{c}\right)}$
(Where $\mathit{s}=\frac{32}{2}=16\mathrm{}\mathit{c}\mathit{m})$
$=\sqrt{16\left(16-11\right)\left(16-13\right)\left(16-8\right)}$
$=\sqrt{16\times 5\times 3\times 8}$
$=8\sqrt{30}\mathrm{}{\mathit{c}\mathit{m}}^2$