Question

A circular garden of radius $10m$ has a straight line fence between the two points on the boundary of the garden. The fence lies inside the garden arena. This fence separates the walking area which is a small region from the plants. The fence is at a distance of $6m$ from the centre of the garden. What is the walking area in $m^2$? It is given that $\cos {53}^{\circ }=\frac{3}{5}$.

• A. 36
• B. 44.5
• C. 12
• D. 72

Answer 1

The correct option is B

44.5

In $△ORQ,O{Q}^2=O{R}^2+R{Q}^2$ [Pythagoras Theorem]

$\Rightarrow {10}^2=6^2+R{Q}^2$

$\Rightarrow R{Q}^2={10}^2-6^2$

$\Rightarrow R{Q}^2=64$
$\Rightarrow RQ=8m$

$\Rightarrow PQ=2RQ=16m$ [OR is perpendicular bisector of PQ]

Area of $△OPQ=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times PQ\times OR$
$=\frac{1}{2}\times 16\times 6$
$=48m^2$

Area of sector OPSQ $=\frac{\angle POQ}{{360}^{\circ }}\times \pi \times r^2$
In $△ORQ,$
$\cos \left(\angle ROQ\right)=\frac{\text{adjacent side}}{\text{hypotenuse}}$
$=\frac{OR}{OQ}$
$=\frac{6}{10}$
$=\frac{3}{5}$
$\Rightarrow \cos \left(\angle ROQ\right)=\frac{3}{5}$
It is given that $\cos {53}^{\circ }=\frac{3}{5}$.
Hence, $\angle ROQ={53}^{\circ }$

$\angle POQ=2\left(\angle ROQ\right)={106}^{\circ }$

Area of sector POQS $=\frac{POQ}{360}\times \pi \times r^2$
$=\frac{{106}^{\circ }}{{360}^{\circ }}\times \pi \times {10}^2=92.5m^2$

Area of segment PRQS = Area of sector POQS – Area of $△OPQ=92.5–48=44.5m^2$