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 area. The fence is at a distance of 6 m 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

${10}^2$ = ${OR}^2$ + ${RQ}^2$

${10}^2$ = $6^2$ +${RQ}^2$

$\Rightarrow$ ${RQ}^2$ = 64

$\Rightarrow$ RQ = 8 cm

$\therefore$ PQ = 2RQ = 16 cm (OR is perpendicular bisector of PQ)

Area of $\Delta$OPQ = $\frac{1}{2}$ × Base × Height

= $\frac{1}{2}$ × PQ × OR

= $\frac{1}{2}$ × 16 × 6

= 48 $c{m}^2$

Area of sector OPSQ = $\frac{\angle POQ}{{360}^{\circ }}$ × $\pi \times r^2$

In ORQ,

cos($\angle$ROQ) = $\frac{adjacentside}{hypotenuse}$

= $\frac{OR}{OQ}$

= $\frac{6}{10}$

= $\frac{3}{5}$

$\Rightarrow$ cos( $\angle$ROQ) = $\frac{3}{5}$

It is given that cos${53}^{\circ }$ = $\frac{3}{5}$.

Hence ROQ = ${53}^{\circ }$

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

Area of sector POQS = $\frac{POQ}{360}$ ×$\pi \times r^2$

= $\frac{{106}^{\circ }}{{360}^{\circ }}$ ×$\pi \times {10}^2$ = 92.5 $m^2$

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