Area of a hollow circle – Solved Examples

Area of a hollow circle

A hollow circle is also known as a circular ring or annulus. It consists of two concentric circles. The radius of these circles are different from each other

In the figure given below, it can be seen that the measure of the region between the two concentric circles is equal to the area of a hollow circle. The radius of the bigger circle and the smaller circle are R_{1} \text { and } R_{2}respectively.

In order to find the area of the circular ring, we have to subtract the area of the smaller circle from the area of the bigger circle.

\text { Area }=\pi\left(R_{1}{ }^{2}-R_{2}{ }^{2}\right)

The area of the hollow circle is actually the area of the base of a hollow cylinder.

Derivation of the formula

We know that, 

Area of the bigger circle having radius \mathrm{R}_{1}=\pi \times \text { radius }^{2}=\pi R_{1}^{2}

Area of the smaller circle having radius \mathrm{R}_{2}=\pi \times \text { radius }^{2}=\pi R_{2}^{2}

Area of the hollow circle = Area of the bigger circle – Area of the smaller circle

                                             = \pi R_{1}^{2}-\pi R_{2}^{2}

                                             = \pi\left(R_{1}^{2}-R_{2}^{2}\right)

                                             = \pi\left(R_{1}+R_{2}\right)\left(R_{1}-R_{2}\right)

 Hence it is proved that the area is equal to \pi\left(R_{1}^{2}-R_{2}^{2}\right)

Solved Examples

1. Find the area of a hollow circle having an inner radius of 11 cm and an outer radius of 19 cm?

Solution:

Outer Radius=R_{1}=19 \mathrm{~cm}

Inner Radius =\mathrm{R}_{2}=11 \mathrm{~cm}

Area =\pi\left(R_{1}{ }^{2}-R_{2}{ }^{2}\right)

=\pi\left(19^{2}-11^{2}\right)=\pi(361-121) =753.98 \mathrm{~cm}^{2}

Hence the area is equal to 753.98 \mathrm{~cm}^{2}.

2. The area of the base of a hollow cylinder is equal to 506 m². If the outer radius of the cylinder is equal to 15 m. Find the inner radius.

Solution:

The base of a cylinder is a hollow circle

\text{ Outer radius} =R_{1}=15 \mathrm{~m}

\text { Area } =\pi\left(R_{1}{ }^{2}-R_{2}{ }^{2}\right)

\Rightarrow 506=\pi\left(15^{2}-R_{2}{ }^{2}\right)

\Rightarrow \mathrm{R}_{2}=8 \mathrm{~m}

The inner radius is equal to 8 m.