Question

PQRS is the diameter of a circle of radius 3 cm. The lengths PQ, QR and RS are equal. Find the area of the shaded portion

• A. $2\pi c{m}^2$
• B. $\frac{5\pi }{2}c{m}^2$
• C. $\frac{9\pi }{2}c{m}^2$
• D. $\frac{11\pi }{2}c{m}^2$

Answer 1

The correct option is B $\frac{5\pi }{2}c{m}^2$

$GiventhatthesemicirclewithdiameterPShasaradius{r}_1=3cm.\therefore PS=3\times 2cm=6cm.AlsoPQ=QR=RS.\therefore PQ=QR=RS=\frac{6}{3}cm=2cm.Soar.semicircleofdia.PSi.eradius{r}_1=3cmis\frac{1}{2}\times \pi \times {r_1}^2=\frac{1}{2}\times \pi \times 3^2{cm}^2=\frac{9}{2}\times \pi {cm}^2.AgainthesemicirclewithdiameterQShasaradius{r}_2=\frac{1}{2}(QR+RS)=\frac{1}{2}(2+2)cm=2cm.Soar.semicircleofdia.QSi.eradius{r}_2=2cmis\frac{1}{2}\times \pi \times {r_2}^2=\frac{1}{2}\times \pi \times 2^2{cm}^2=2\pi {cm}^2.Nowar.shadedportion=ar.semicircleofdia.PS-ar.semicircleofdia.QS=\frac{9}{2}\times \pi {cm}^2-2\pi {cm}^2=\frac{5\pi }{2}{cm}^2.Ans-ar.shadedportion=\frac{5\pi }{2}{cm}^2.$