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Subsection 2.6.1 Area of an Annulus
Activity 2.6.1 .
\(\textbf{Work in groups }\)
What you require: Circular objects (e.g., two different-sized cups, lids, or rings), ruler or measuring tape and pen and paper (or a calculator).
Look around your surroundings and find two circular objects that can fit inside each other (e.g., two different-sized bowls, two bottle caps, or two CDs).
Place the smaller object inside the larger one to visualize the annulus (ring shape).
Measure the radius of the larger circle (
\(R\) ) from its center to the edge.
Measure the radius of the smaller circle (
\(r\) ) in the same way.
Square both radii. and record your result.
Subtract the squared radius of the smaller circle from the squared radius of the larger circle.
Multiply the result by
\(\frac{22}{7}\) or
\(3.142\)
Discuss with your group how to calculate the area of an annulus.
Try this activity with different circular objects and compare your results.
\(\textbf{Extended Activity}\)
Activity 2.6.2 .
\(\textbf{Individual work}\)
Situation: Imagine a running track built around a circular field. The track has an inner boundary (smaller circle) and an outer boundary (larger circle). The track itself forms an annulus.
A school is planning to paint the running track.
The inner radius of the track is
\(30\) meters, and the outer radius is
\(35\) meters.
The cost of painting is Ksh
\(5\) per square meter
The total cost of painting the track.
\(\textbf{Key Takeaway}\)
An
\(\textbf{annulus}\) is the region between two concentric circles that share the same center but have different radii as shown below.
The area of an annulus (a ring-shaped object) is found by subtracting the area of the smaller, inner circle from the area of the larger, outer circle.
\begin{align*}
A_\text{annulus} = \amp A_\text{outer circle} - A_\text{inner circle} \\
= \amp \pi R^2 - \pi r^2 \\
= \amp \pi (R^2 - r^2)
\end{align*}
Example 2.6.1 .
Find the area of an the annulus drawn below;
Solution .
\(R=10\,cm\) and
\(r=6\,cm\)
\begin{align*}
A= \amp \pi(R^2-r^2) \\
=\amp \pi(10^2-6^2) \\
=\amp \pi(100-36)\\
=\amp 64\pi \\
=\amp 64 \times \frac{22}{7}\\
= \amp 201.06 \,cm^2
\end{align*}
Example 2.6.2 .
A wheel has an outer radius of
\(40 \,cm\text{,}\) and its inner hub has a radius of
\(10 \,cm\text{.}\) Find the area of the wheel’s annular region.
Solution .
The outer radius of the wheel
\(= 40\,cm\text{.}\)
Inner hub radius
\(=10\,cm\)
\begin{align*}
A= \amp \pi(R^2-r^2)\\
=\amp \frac{22}{7}(40^2-10^2) \\
=\amp \frac{22}{7}(1\,600-100)\\
=\amp \frac{22}{7} \times 1\,500 \\
=\amp 4\,712.39 \,cm^2
\end{align*}
A ring-shaped garden has an outer radius of
\(12\) meters and an inner radius of
\(7\) meters. Find the area of the garden.
A circular tabletop has a hole in the middle for an umbrella. The outer radius of the table is
\(1.5 \,m\text{,}\) and the hole has a radius of
\(0.5 m\) as shown below.
Find the area of the tabletop.
A circular swimming pool has an outer radius of
\(8\) meters, and a smaller circular island is in the center with a radius of
\(2\) meters. Find the area of the water surface.
A steel pipe has an outer diameter of
\(80\) units and an inner diameter of
\(60\) units, what is the area of the cross-section?
Find the area of the figure below; use
\(\pi=3.142\)
What is the area of the annulus shown; (Leave your answer to
\(3\) significant figures).
Checkpoint 2.6.3 .
Checkpoint 2.6.4 .
