Question

In the given figure , AC is diameter of a circle with radius 5cm. If AB = BC and CD = 8 cm, the area of the shaded region to the nearest whole number is

• A. $28c{m}^2$
• B. $29c{m}^2$
• C. $30c{m}^2$
• D. $45c{m}^2$

Answer 1

The correct option is C $30c{m}^2$

Given that, AC is the diameter of a circle with radius $5cm$
and $AB=BC,CD=8cm$

$\therefore AC=10cm$
Area of the circle $=\pi (5{)}^2$$=25\pi c{m}^2$ $[\because \text{Area of a circle}=\pi r^2\text{square units}]$
We know that, angle subtended by the diameter of a circle at any point on the circle is a right angle.

Hence, $△ABC$ is a right angled triangle with $\angle B={90}^o$
$\therefore A{C}^2=A{B}^2+B{C}^2$ [Pythagoras theorem]
$\Rightarrow 100=2A{B}^2$
$\Rightarrow A{B}^2=50$
$\Rightarrow AB=\sqrt{50}$
$\Rightarrow AB=5\sqrt{2}$
$\therefore AB=BC=5\sqrt{2}$
Area of $△ABC$$=\frac{1}{2}\times AB\times BC$

$\Rightarrow \frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}$$=25c{m}^2$
Similarly,
$△ADC$ is a right angled triangle with $\angle D={90}^o$
$\therefore A{C}^2=A{D}^2+D{C}^2$
$\Rightarrow {10}^2=A{D}^2+8^2$
$\Rightarrow 100=A{D}^2+64$
$\Rightarrow A{D}^2=36$
$\Rightarrow AD=6$
Area of $△ADC$$=\frac{1}{2}\times AD\times DC$

$\Rightarrow \frac{1}{2}\times 6\times 8$$=24c{m}^2$
Area of shaded region $=25\pi -25-24$

$\Rightarrow 25\pi -49=29.5c{m}^2$

The nearest whole number of $29.5$ is $30$.

Hence, the area of the shaded region to the nearest whole number is $30c{m}^2$