Question

$△PQR$ is inscribed in a semicircle. $PQ=PR=7$ cm.
(i) What is the measure of $\angle QPR?$
(ii) Find the area of the shaded region.

Answer 1

$PQ=PR=7㎝$

$OP=OR$

${OP}^2+{OR}^2={PR}^2$

$\sqrt{x^2+x^2}=PR\Rightarrow PR=\sqrt{2}x$

$OR=\frac{7}{\sqrt{2}}\sqrt{2}x=7;x=\frac{7}{\sqrt{2}}$

$QR=\frac{7}{\sqrt{2}}\times 2=7\sqrt{2}$

$\angle QPR=\cos \theta =\frac{{QP}^2+{PR}^2-{QR}^2}{2\times QP\times PR}$

$=\frac{7^2+7^2-{\left(7\sqrt{2}\right)}^2}{2\times 7\times 7}=\frac{49+49-98}{2\times 49}$

$\cos \theta =0$

$\therefore \theta =90$

$\angle QPR={90}^{\circ }$

Area of shaded region=area of semicircle-area of triangle

$=\frac{\pi x^2}{2}-\frac{1}{2}\times 7\times 7$

$=(\pi \times {\left(\frac{7}{\sqrt{2}}\right)}^2-49)\frac{1}{2}$

$=\frac{49}{2}(\frac{\pi }{2}-1)-(1.57-1)\frac{49}{2}$

$=0.57\times 24.5$

$=13.965$sq.cm