Question

In the figure QR is a tangent to the circle with centre O. Point Q is the point of contact. Radius of the circle is 10 cm. $OR=20$ cm. Find the area of the shaded region.($\pi$ = 3.14,$\sqrt{3}$= 1.73)

• A. $34.17$ $c{m}^2$
• B. $44.17$ $c{m}^2$
• C. $54.17$ $c{m}^2$
• D. None of these

Answer 1

Answer: (A)

$34.17$ $c{m}^2$

Since line drawn from the centre of a circle to the tangent is perpendicular to the tangent.

$\therefore OQ$ perpendicular to $QR$

$△OQR$ is a right angled triangle.

$\therefore O{R}^2=Q{R}^2+O{Q}^2$

$\Rightarrow QR=\sqrt{O{R}^2-O{Q}^2}$

$=\sqrt{{20}^2-{10}^2}$

$=\sqrt{400-100}$

$=\sqrt{300}$

$=10\sqrt{3}$

$\therefore$ Area of $△$ OQR=$\frac{1}{2}\times QR\times OQ$

=$\frac{1}{2}\times 10\sqrt{3}\times 10$

=$86.5c{m}^2$

Also,let $\angle QOR=\theta$

$\sin \theta =\frac{QR}{OR}$

$=\frac{10\sqrt{3}}{20}$

$\theta ={60}^o$

$\therefore$ Area of sector OQT=$\pi (10{)}^2\times \frac{60}{360}$

=$\frac{22}{7}\times 100\times \frac{60}{360}$

=$52.33c{m}^2$

Area of the shaded region = Area of $△$ $OQR$ - Area of sector $OQT$

= $(86.5-52.33)c{m}^2$

= $34.17c{m}^2$