SSC CGL Tier II 21 Sep 2014 Paper

(3) Side of a rhombus $=\frac{2\mathit{p}}{4}=\frac{\mathit{p}}{2}$ units
OA = OC = y (let)
∴ AC = 2y units
OB = OD = x (let)
∴ BD = 2x units
From Δ OAB,
∠AOB = 90°
$\mathit{A}{\mathit{B}}^2=\mathit{O}{\mathit{A}}^2+\mathit{O}{\mathit{B}}^2$
$\Rightarrow \frac{{\mathit{p}}^2}{4}={\mathit{x}}^2+{\mathit{y}}^2$
$\Rightarrow {\mathit{p}}^2=4{\mathit{x}}^2+4{\mathit{y}}^2...\left(\mathit{i}\right)$
and $2\mathit{x}+2\mathit{y}=\mathit{m}$
On squaring both sides,
$4{\mathit{x}}^2+4{\mathit{y}}^2+8\mathit{x}\mathit{y}={\mathit{m}}^2$
$\Rightarrow {\mathit{p}}^2+8\mathit{x}\mathit{y}={\mathit{m}}^2$
$\Rightarrow 8\mathit{x}\mathit{y}={\mathit{m}}^2-{\mathit{p}}^2$
$\Rightarrow 4\mathit{x}\mathit{y}=\frac{1}{2}\left({\mathit{m}}^2-{\mathit{p}}^2\right)$
∴ Area of the rhombus $=\frac{1}{2}\times \mathit{A}\mathit{C}\times \mathit{B}\mathit{D}$
$=\frac{1}{2}\times 2\mathit{x}\times 2\mathit{y}=\frac{1}{2}\times 4\mathit{x}\mathit{y}$
$=\frac{1}{2}\times \frac{1}{2}\left({\mathit{m}}^2-{\mathit{p}}^2\right)$
$=\frac{1}{4}\left({\mathit{m}}^2-{\mathit{p}}^2\right)$ sq. units