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In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and $\angle BOD =90^{\circ}$ . Find the area of the shaded region.

Answers (1)

Given $\rightarrow$

$AC=24\; cm$

$AB=7\; cm$

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

We know BC is diameter of the circle because it is passing through centre.

Theorem : Angle subtended by diameter on the circumference of the circle is $90^{\circ}$ .

Hence $\angle B\!AC =90^{\circ}$

In $\bigtriangleup ABC$ with $\angle B\!AC=90^{\circ}$

$BC^2=AB^2+AC^2$ (Pythagoras theorem)

$BC=\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}$

$BC=25\; cm$

$\text{Radius of circle}=\frac{BC}{2}=\frac{25}{2}=12.5$

Shaded Area = Area of semicircle - Area of $\bigtriangleup ABC$ + area of sector DOB

$=\frac{\pi }{2}r^2-\frac{1}{2}\times AB \times AC + \frac{90^{\circ}}{360^{\circ}}\pi r^2$

$=\frac{\pi }{2}\times (12.5)^2-\frac{1}{2}\times 24 \times 7 + \frac{1}{4}\pi (12.5)^2$

$=\pi (12.5)^2\left ( \frac{1}{2}+\frac{1}{4} \right )-12\times 7$

$=\frac{22}{7}\times (12.5)^2\times \frac{3}{4}-12\times 7$

$=\frac{22}{7}\times 156.25 \times \frac{3}{4}- 84$

$=\frac{10312.5}{28}- 84$

$=368.3 - 84$

$=284.3\: cm^2$

Posted by

Ravindra Pindel

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