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Grade 10 Mathematics Learners’ Book

INNODEMS

Subsection 2.6.4 Area of a Segment of a Circle

Activity 2.6.6.

\(\textbf{Work in groups}\)
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What you require;
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  1. Use a compass to draw a circle of radius \(r\text{.}\)
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  2. Draw a chord across the circle using a ruler.
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  3. Use a protractor to measure the angle subtended at the center by the chord.
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    Shade the segment formed
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  4. Find the area of the segment part and record your results.
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  5. Perform the same process for different chords and angles eg \(70^\circ, \,90^\circ,\,120^\circ,\,150^\circ...\)
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  6. Discuss your work with other learners in your class.
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\(\textbf{Key Takeaway}\)
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A \(\textbf{segment} \) is the region of a circle bounded by a chord and an arc as shown in the figure below.
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The shaded region is a segment of the circle with centre \(O\) and the radius \(q\)
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\(\textbf{Area of a segment of a circle.}\)
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The area of a segment is given by;
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\(\text{Area of a segment}=\text{Area of sector}-\text{Area of triangle}\)
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Example 2.6.14.

A chord in a circle of radius \(7\, cm\) subtends an angle of \(75^\circ\) at the center as shown below. Find the area of the segment.
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Solution.
To find the area of the segment, you the area of the triangle and subtract from the area of the sector.
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\(\textbf{Area of sector }\)
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\(A= \frac{\theta}{ 360} \times \pi r^2\)
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\begin{align*} A=\amp \frac{75^\circ}{ 360} \times \frac{22}{7} \times 7^2 \\ =\amp \frac{5}{24} \times \frac{22}{7} \times 49\\ =\amp \frac{385}{12} \\ =\amp 32.0833 \,cm^2 \end{align*}
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\(\textbf{Area of a triangle }\)
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\(A= \frac{1}{2} ab sin \,\theta\)
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Where \(a=7\,cm \,\text{and} \,b= 7 \,cm\)
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\begin{align*} A=\amp \\ =\amp \frac{1}{2} \times 7 \times 7\times \times sin \,75^\circ\\ =\amp \frac{1}{2} \times 49 \times sin \,75^\circ\\ =\amp 23.6652 \,cm^2 \end{align*}
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Therefor,
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\begin{align*} A= \amp 32.0833 \,cm^2- 23.6652 \,cm^2 \\ = \amp 8.4181\,cm^2 \end{align*}
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Example 2.6.15.

A chord XY of length \(14 \,cm\) is drawn in a circle with centre O and radius \(8 cm\text{,}\) as in the figure below
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Calculate:
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  1. the distance ON.
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  2. the area of the sector OXPY.
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  3. the area of triangle OXY
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  4. the area of the minor segment.
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  5. the area of the major segment.
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Solution.
Given a circle with center \(O\text{,}\) radius \(8\) cm, and a chord \(XY\) of length \(14\,cm\) , we find the required values step by step.
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  1. Finding the Distance \(\text{ON}\)
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    Since ON is the perpendicular bisector of XY, we use the right triangle ONX where:
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    \begin{align*} \text{OX}=\amp 8\,cm (\text{radius}) \\ \text{NX}=\amp \frac{\text{XY}}{2} \\ = \amp \frac{12}{2} \\ = \amp 7\,cm \end{align*}
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    Using Pythagoras’ Theorem in \(\triangle\) ONX:
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    \begin{align*} \text{ON}^2+ \text{NX}^2 =\amp \text{OX}^2\\ \text{ON}^2+ 7^2= \amp 8^2 \\ \text{ON}^2+ 49= \amp 64 \\ \text{ON}^2=\amp 64-49\\ \text{ON}^2= \amp 15 \\ \text{ON} =\amp \sqrt{15} \\ ​≈\amp 3.87 \,cm \end{align*}
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  2. Finding the area of sector OXPY.
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    Finding the \(\angle\) XOY which is given by:
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    \begin{align*} cos\,\theta= \amp (\frac{\text{XN}}{\text{XO}} )\\ =\amp \frac{7}{8} \\ =\amp 0.875 \\ \theta=\amp cos^{-1}(0.875) \\ =\amp 28.9550^\circ \end{align*}
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    therefor, \(\angle \text{XOY}= 28.9550^\circ\times 2\)
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    \begin{equation*} =57.9100^\circ \end{equation*}
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    The area of a sector is:
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    \begin{align*} A=\amp \frac{\theta}{360} \times \pi r^2\\ =\amp \frac{57.9100}{360} \times \pi 8^2 \\ =\amp \frac{57.9100}{360} \times \frac{22}{7} \times 64\\ =\amp 32.3561 \,cm^2 \end{align*}
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    Therefore, area of sector OXPY \(= 32.3561 \,cm^2\)
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  3. Finding the area of triangle OXY
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    \begin{align*} \text{Area}= \amp \frac{1}{2} \times \text{base} \times \text{height} \\ =\amp \frac{1}{2} \times 14 \times 3.87 \,cm \\ = \amp 7 \times 3.87 \,cm \\ =\amp 27.09 \,cm^2 \end{align*}
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  4. Finding the area of the minor segment.
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    \(\text{Area of minor segment}=\text{Area of sector}- \text{Area of} \triangle \text{OXY}\)
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    \begin{align*} \text{Area of minor segment}=\amp \\ =\amp 32.3561 \,cm^2-27.09 \,cm^2 \\ =\amp 5.2661\,cm^2 \end{align*}
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  5. Finding the area of the major segment
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    Area of major segment \(=\) Total circle area\(-\)Area of minor segment
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    Area of a circle.
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    \begin{align*} A= \amp \pi r^2 \\ =\amp \frac{22}{7} \times 8^2 \\ =\amp \frac{22}{7} \times 64 \\ =\amp 201.1429 \,cm^2 \end{align*}
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    Therefore, Area of major segment is
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    \begin{align*} = \amp 201.1429 \,cm^2- 5.2661\,cm^2\\ =\amp 195.8768\,cm^2 \end{align*}
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\(\textbf{Exercises}\)
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  1. In the figure below, ADC is a chord of a circle with centre O passing through A, B and C. BD is a perpendicular bisector of AC. AD \(= 4 \,cm\) and BD \(= 2 \,cm\text{.}\)
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    Calculate:
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    1. the radius OA of the circle.
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    2. the area of the sector OABC.
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    3. the area of the segment ABCD
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  2. A circular table has a radius of \(15 \,cm\text{.}\) A slice of cake is cut out, forming a \(75^\circ \) segment. Find the area of the cake slice not covered by the straight cut
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  3. A wheel of a car has a radius of \(30 \,cm\text{.}\) A mudguard covers a \(60^\circ\) segment of the wheel. Find the area of the covered segment.
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  4. A circular park has a radius of \(20\) meters. A walking path cuts across the park, forming a chord that subtends an angle of \(120^\circ\) at the center as shown below.
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  5. A chord XY subtends an angle of \(120^\circ\) at the centre of a circle of radius \(13\, cm\text{.}\) Calculate the area of the minor segment.
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Checkpoint 2.6.16.

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Checkpoint 2.6.17.

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