Question

In Fig. $AB$ is a diameter of the circle, $AC$ $=6cm$ and $BC$ $=8cm$. Find the area of the shaded region (Use $\pi =3.14$).

• A. $54.5$ $c{m}^2$
• B. $290$ $c{m}^2$
• C. $281$ $c{m}^2$
• D. $78.5$ $c{m}^2$

Answer 1

The correct option is A $54.5$ $c{m}^2$

Given:

$AB$ is the diameter $=d$ of a circle.

$\Delta ABC$ has the diameter $AB$ as base & the point $C$ is on the circumference.

$AC=6cm$ and $BC=8cm.$

To find out:

Area of shaded portion in the given circle.

Solution:

$\angle ACB={90}^o$ since $\Delta ABC$ has been inscribed in a semicircle.

$\therefore \Delta ABC$ is a right one with $AB$ as hypotenuse ...(i)

So, applying Pythagoras theorem, we have

$AB=\sqrt{{\left(AC\right)}^2+{\left(BC\right)}^2}=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}cm=10cm=d.$

$\therefore$ The radius of the given circle $=\frac{d}{2}=\frac{10}{2}cm=5cm.$

i.e The Area of circle $=\pi r^2=3.14\times 5^2{cm}^2=78.5{cm}^2.$

Again, Area of $\Delta ABC=\frac{1}{2}\times AC\times BC$ (by i)

$=\frac{1}{2}\times 6\times 8{cm}^2=24{cm}^2.$

Now, Area of shaded region $=$ Area of circle $-$ area of $\Delta ABC$

$=(78.5-24){cm}^2=54.5{cm}^2.$