TS ICET 23 May 2018 Shift 2 Paper

The side of the rhombus = 68/4 = 17 m
Since the diagonals of the rhombus bisect each other perpendicularly, ∆ AOD is right-angled.
$\mathit{O}{\mathit{D}}^2=\mathit{A}{\mathit{D}}^2-\mathit{A}{\mathit{O}}^2$
$\Rightarrow \mathit{O}\mathit{D}=\sqrt{{17}^2-8^2}=15$
The other diagonal BD$=2\times 15=30$
Area of the rhombus
$=\frac{1}{2}\times {\mathit{d}}_1{\mathit{d}}_2=\frac{1}{2}\times 16\times 30$
$=240{\mathit{m}}^2$