Question

In the figure, PR and QS are two diameters of the circle.
If $PR=28cm$ and $PS=14\sqrt{3}$ cm, then find the total area of two shaded segments. ($\sqrt{3}$ = $1.73$)

• A. $249.12$ $c{m}^2$
• B. $240.76$ $c{m}^2$
• C. $251.12$ $c{m}^2$
• D. None of these

Answer 1

The correct option is B $240.76$ $c{m}^2$

In the given figure,

Drop a perpendicular from $O$ to side $QR$.

It is given that $PR=28cm$

$\therefore OR=\frac{PR}{2}=\frac{28}{2}=14cm$

So, the radius of circle is $14cm$

Also,

$PS=QR=14\sqrt{3}$

Now, after construction, in$△RTO$

$\angle OTR={90}^0,TR=\frac{14\sqrt{3}}{2}=7\sqrt{3},OR=14cm$

Using pythagorus,

$OT=\sqrt{{14}^2-{7\sqrt{3}}^2}=\sqrt{196-147}=7cm$

Area of triangle $ROQ$

$=\frac{1}{2}\times OT\times QR=\frac{1}{2}\times 7\times 14\sqrt{3}{cm}^2$

$=49\sqrt{3}{cm}^2$

Area enclosed by of segment $QR$ and minor arc $QR=$

Area of minor sector $ROQ-$ area of $\Delta ROQ$

$=\frac{120}{360}\times \pi (14{)}^2-49\sqrt{3}$

$=205.25-84.87=120.38{cm}^2$

Area of shaded region $=2\times ($Area of minor sector $ROQ)$

$=2\times 120.38=240.76{cm}^2$