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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.5.3 Area of Regular Polygons

A Regular Polygon is a shape with all sides and angles equal. Examples include equilateral triangles, squares, regular pentagons, hexagons, and so on.
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Below is a visual representation of regular polygons from a triangle to an octagon. These diagrams are displayed in a grid format for better visualization.
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The \(\textbf{area}\) of a \(\textbf{regular polygon}\) can be found using this formula, where \(\textbf{P}\) is the \(\textbf{perimeter}\) and \(\textbf{a}\) is the apothem
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The \(\textbf{apothem}\) is the distance from the center of a regular polygon to \(\, \textbf{a} \)side. Where \(\textbf{a}\) is the half \(\frac{1}{2} \text{side.}\)
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You want to find the area of the triangle. You can see it has an apothem of \(\sqrt{3}\) meters and a side length of 10 meters.
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If you also find the perimeter, P , then you can use the formula A= \(\frac{1}{2} \textbf{P}a\) to find the area.
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\begin{equation*} \textbf{P} = \text{length of each side } \times a \text{, where } a = h \end{equation*}
\begin{equation*} P = 10 \, \text{m} \times 3 \end{equation*}
\begin{equation*} = 30, \text{m} \end{equation*}
Now, use the formula for the area of a regular polygon is :
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\begin{align*} A = \amp \frac{1}{2} \times 30 \times \sqrt{3} \\ = \amp 125.98 \, \text{m}^2 \end{align*}
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Example 2.5.34.

A regular hexagon with center h is shown below
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Similarly you can use \(A = \frac{1}{2} \times n \times S \times a\)
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Solution.
Find the perimeter P, Then use the formula A= \(\frac{1}{2} \textbf{P}a\) to find the area.
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\begin{equation*} \textbf{P} = \text{length of each side } \times a \end{equation*}
\begin{equation*} P = 20 \, \text{cm} \times 6 \end{equation*}
\begin{equation*} = 120 \, \text{cm}^2 \end{equation*}
Now, use the formula for the area of a regular polygon is :
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\begin{align*} A = \amp \frac{1}{2} \times 120 \times 14 \, \text{cm} \\ = \amp 840 \text{cm}^2 \end{align*}
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Where \(n = \text{number of sides,} S = \text{length of each side and } a = \frac{S}{2} \)
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\begin{align*} A = \amp \frac{1}{2} n \times S \times a\\ = \amp \frac{1}{2} \times 6 \times 20 \, \text{cm} \times 14 \, \text{cm}\\ = \amp \frac{1,680}{2} \, \text{cm}^2\\ = \amp 840 \,\text{cm}^2 \end{align*}
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Subsubsection 2.5.3.1 Area of Heptagons

A heptagon is a seven-sided polygon. It has seven edges and seven vertices.
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Learner Experience 2.5.10.

Constructing a Regular Heptagon
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Working in groups of 3-5 students.
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\(\textbf{Materials needed:}\)
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  1. Draw a Circle
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    • Place the compass pointer on the paper and draw a circle of any radius.
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    • Mark the center (O) of the circle.
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  2. Draw a Horizontal Diameter
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    • Use the ruler to draw a straight line through the center (O), creating a diameter.
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    • Label the first point P1 on the circumference.
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  3. Divide the Circle into Seven Equal Parts
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    • Use a protractor to measure angles of \(\frac{360^\circ}{7} \approx 51.43^\circ\) from point P1.
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    • Mark each point on the circumference as P2, P3, P4, and so on, until you have seven points.
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  4. Connect the Points
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    • Using the ruler, connect the points P1, P2, P3, P4, P5, P6, and P7 to form the heptagon.
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Key Takeaway 2.5.36.

\(\textbf{Properties of a Regular Heptagon.}\text{.}\)
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  • Sum of interior angles = \((n-2) \times 180^\circ\text{.}\) Where n is the number of sides.
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  • Example for a Heptagon (7-sided polygon).
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  • \begin{equation*} (n-2) \times 180^\circ \end{equation*}
    \begin{equation*} (7-2) \times 180^\circ = 900^\circ \end{equation*}
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  • Sum of exterior angles of any \(\textbf{polygon(regular or irregular)}\) is always equal to \(360^\circ\) for both regular and irregular polygons.
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  • Each Interior and Exterior Angle (Regular Polygon Only)
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    \begin{equation*} \textbf{Each Interior Angle (for a regular polygon):} \end{equation*}
    \begin{equation*} Interior Angle = \frac{(n-2) \times 180^\circ}{n} \end{equation*}
    \begin{equation*} \textbf{Each Exterior Angle (for a regular polygon):} \end{equation*}
    \begin{equation*} Exterior Angle = \frac{360^ \circ}{n} \end{equation*}
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  • Example for a Regular Heptagon:
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    \begin{equation*} \textbf{Each Interior Angle} = \frac{900}{7} \end{equation*}
    \begin{equation*} = 128.57^ \circ \end{equation*}
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    \begin{equation*} \textbf{Each Exterior Angle} = \frac{360}{7} \end{equation*}
    \begin{equation*} =51.43^ \circ \end{equation*}
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\(\textbf{Study Questions}\)
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  1. What is the sum of all interior angles of a heptagon?
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  2. How do we calculate the measure of one interior angle of a regular heptagon?
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  3. Can you find heptagons in real life?
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Example 2.5.37.

A regular heptagon measures 10 cm as indicated in the figure below. Find its area. Use \(\sin 51.43^\circ = 0.7818\text{.}\)
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Solution.
There are 7 triangles since a heptagon has 7 sides. Area of Triangle \(P_0OP_1 = \frac{1}{2} ab \, \sin \, 51.43^\circ \)
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\begin{align*} \amp \triangle P_0OP_1 = \frac{1}{2}\times 10\times 10\, \sin \,51.43^\circ \\ \amp = \frac{1}{2}\times 100\, \text{cm}^2 \times \sin \, 51,43^\circ\\ \amp = \frac{1}{2}\times 100\, \text{cm}^2 \times 0.7818 \\ \amp = 39.0923 \, \text{cm}^2 \end{align*}
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Therefore the total area
\begin{align*} \amp = 7 \times 39.0923 \, \text{cm}^2 \\ \amp = 273.65 \,\text{cm}^2 \end{align*}
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Example 2.5.38.

Find the area of the regular heptagon given that its side measures \(20\) cm.
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Solution.
  1. Sketch the heptagon
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  2. Find the interior angle of the heptagon
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    Let the interior angle be \(\theta\text{,}\) we can calculate it by
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    \(\theta = \frac{900}{7} \approx 128.57 ^\circ\)
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  3. Calculate the angle formed at the centre by the triangle extracted
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    \(\beta = \frac{360}{7} \approx 51.43^\circ\)
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  4. Find the area of the triangle formed. Notice that the base angles of the triangle formed is half the interior angle \(\theta\)
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    \begin{align*} \text{Area} = \amp \frac{1}{2} \times \text{Base} \times \text{Height}\\ = \amp \frac{1}{2} \times 20 \times 10 \times \tan 64.285^\circ\\ = \amp 100 \tan 64.285^\circ \\ = \amp 207.65 \; \text{cm}^2 \end{align*}
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  5. The regular heptagon have \(7\) equal triangles, therefore its area is given by
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    \(7 \times 207.65 \; \text{cm}^2 = 1453.55 \; \text{cm}^2\)
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Exercises Exercises

1.
A regular heptagon has a side length of \(6 \text{cm}\text{.}\) Calculate its perimeter.
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Solution.
Perimeter = \(7 \times 6 = 42 \text{cm}\text{.}\)
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2.
A regular heptagon has a radius (distance from the center to a vertex) of \(5 \text{cm}\text{.}\) Draw the heptagon and calculate its area. Use \(\sin 51.43^\circ = 0.7818\text{.}\)
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Solution.
Area of one triangle = \(\frac{1}{2} \times 5 \times 5 \times \sin 51.43^\circ = 9.77 \text{cm}^2\text{.}\)
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Total area = \(7 \times 9.77 = 68.39 \text{cm}^2\text{.}\)
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3.
A regular heptagon has a side length of \(4 \text{cm}\text{.}\) Draw the heptagon and calculate its area by dividing it into seven isosceles triangles and finding the area of one triangle. Use \(\sin(\frac{360^\circ}{7}) \approx 0.4339\text{.}\)
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Solution.
Area of one triangle = \(\frac{1}{2} \times 4 \times 4 \times \sin(\frac{360^\circ}{7}) \approx 6.93 \text{ cm}^2\text{.}\)
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Total area = \(7 \times 6.93 \approx 48.51 \text{ cm}^2\text{.}\)
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4.
A regular heptagon is inscribed in a circle with a radius of \(8 \text{cm}\text{.}\) Calculate the area of the heptagon. Use \(\sin 51.43^\circ = 0.7818\text{.}\)
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Solution.
Area of the circle = \(\pi r^2 = \pi \times 8^2 = 64\pi \text{ cm}^2\text{.}\)
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Area of one triangle = \(\frac{1}{2} \times 8 \times 8 \times \sin 51.43^\circ = 24.9792 \text{ cm}^2\text{,}\) so the heptagon area = \(7 \times 24.9792 = 174.8544 \text{ cm}^2\text{.}\)
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Fraction of the circle occupied by the heptagon:
\begin{gather*} \dfrac{\text{Area of heptagon}}{\text{Area of circle}} = \dfrac{174.8544}{64\pi} \approx 0.8695 \end{gather*}
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Exact expression (in terms of the central angle):
\begin{gather*} \dfrac{\text{Area of heptagon}}{\text{Area of circle}} = \dfrac{7\left(\tfrac{1}{2}r^2\sin\tfrac{360^\circ}{7}\right)}{\pi r^2} = \dfrac{7}{2\pi}\sin\left(\tfrac{360^\circ}{7}\right) \end{gather*}
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As a percentage: \(0.8695 \times 100\% \approx 86.95\%\text{.}\)
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Subsubsection 2.5.3.2 Area of Octagons

A regular octagon is an 8-sided polygon with eight vertices (corners) and ten edges (sides).
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Learner Experience 2.5.11.

Constructing a Regular Octagon
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Working in groups of 3-5 students.
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\(\textbf{Materials needed:}\)
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  1. Draw a Circle
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    • Use a compass to draw a circle of your desired radius.
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    • Mark the center (O).
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  2. Draw the First Diameter
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    • Use a ruler to draw a horizontal diameter (P0 to P5) passing through the center.
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  3. Divide the Circle into 8 Equal Parts
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    • Use a protractor to measure angles of \(\frac{360^\circ}{8} = 45^\circ\) from point P0.
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    • Mark each 45° interval on the circle to get 8 points.
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  4. Connect the Points
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    • Use a ruler to connect the eight points in order to form the octagon.
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\(\textbf{Discussion Questions}\)
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\(\textbf{Study Questions}\)
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1. What is the sum of all interior angles of an octagon ?
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2. How do we calculate the measure of one interior angle of a regular octagon?
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3. Have you ever seen octagon-shaped objects in real life?
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Key Takeaway 2.5.40.

1. Regular octagons (\(8\)-sided polygons with equal sides and angles) appear in many real-life contexts:
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Architecture: Octagonal pavilions, gazebos, towers
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Landscaping: Flower beds, garden features, paving patterns
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Traffic: Stop signs (regular octagon shape)
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Sports: Octagonal boxing rings, martial arts mats
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Design: Tiles, decorative patterns, building layouts
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Strategy: Divide into Triangles
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The key strategy for finding the area of any regular polygon is to divide it into congruent triangles:
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Draw lines from the center to each vertex
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For an octagon, this creates \(8\) identical triangles
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Each triangle has two sides equal to the circumradius \(R\) and a vertex angle of \(45^\circ\)
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Formula for Area of One Triangle
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Each triangle in the octagon has:
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Two equal sides of length \(R\) (the circumradius)
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Vertex angle \(\theta = 45 ^\circ\)
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Using the formula for the area of a triangle with two sides and an included angle:
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A triangle \(=\frac{1}{2} \times R^2 \times sin(45^\circ)\)
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Since \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\text{,}\) we can simplify:
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A triangle \(=\frac{1}{2} \times R^2 \times \frac{\sqrt{2}}{2} = \frac{R^2 \sqrt{2}}{4}\)
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Formula for Area of Regular Octagon
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Since the octagon contains \(8\) identical triangles:
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A octagon \(= 8 \times \frac{R^2 \sqrt{2}}{4} = 2R^2\sqrt{2}\)
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Therefore, the area of a regular octagon can be expressed in terms of its circumradius \(R\) as:
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\(A = 2R^2\sqrt{2}\)
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Understanding Circumradius
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The circumradius R is the distance from the center of the octagon to any vertex. It is the radius of the circle that passes through all vertices of the octagon (the circumscribed circle).
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Why This Formula Works
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The octagon is divided into \(8\) congruent triangles
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Each triangle has area \(\frac{1}{2}R^2 \sin(45^\circ)\)
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Total area \(=8\times(\frac{1}{2}R^2\sin(45\circ)) = 4R^2\sin(45\circ) = 2R^2\sqrt{2}\)
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This method works for any regular polygon: divide by \(n\text{,}\) use angle \(360^\circ/n\)
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Example 2.5.41.

A regular octagon with O as its centre. If OA is 8 cm, find its area.
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Solution.
There are 8 triangles since an octagon has 8 sides. Area of Triangle \(AOB = \frac{1}{2} ab \text{sin} 45^\circ \)
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\begin{align*} \amp \triangle AOB = \frac{1}{2}\times 8\times 8\, \text{sin} \, 45^\circ \\ \amp = \frac{1}{2}\times 64\, \text{cm}^2 \times \text{sin} \, 45^\circ\\ \amp = \frac{1}{2}\times 64\, \text{cm}^2 \times 0.7071 \\ \amp = 22.6274 \, \text{cm}^2 \end{align*}
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Therefore the total area
\begin{align*} \amp = 8 \times 22.6274 \, \text{cm}^2 \\ \amp = 181.02 \, \text{cm}^2 \end{align*}
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Example 2.5.42.

The figure below represents a regular octagon with O as its centre. If OA is 7cm, find its area
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Solution.
There are 8 triangles since an octagon has 8 sides and the sum of interior angles of one triangle is \(360 ^\circ\text{.}\) Area of Triangle \(AOB = \frac{1}{2} ab \, \text{sin} \, 45^\circ \) .
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The angle of \(\triangle AOB = \frac{360}{8} = 45^\circ \)
\begin{align*} \amp \triangle AOB = \frac{1}{2}\times 7\times 7 \, \, \text{ sin} \, 45^\circ \\ \amp = \frac{1}{2}\times 49\, \text{cm}^2 \times \text{sin}\, 45^\circ\\ \amp = \frac{1}{2}\times 49\, \text{cm}^2 \times 0.7071 \\ \amp = 17.3241 \, \text{cm}^2 \end{align*}
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Therefore the total area
\begin{align*} \amp = 8 \times 17.3241 \, \text{cm}^2 \\ \amp = 138.59\, \text{cm}^2 \end{align*}
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Exercises Exercises

1.
A regular octagon has a side length of \(5 \text{ cm}\text{.}\) Calculate its perimeter.
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Solution.
Perimeter = \(8 \times 5 = 40 \text{ cm}\text{.}\)
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2.
A regular octagon has a radius (distance from the center to a vertex) of \(6 \text{ cm}\text{.}\) Draw the octagon and calculate its area. Use \(\text{sin } 45^\circ = 0.7071\text{.}\)
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Solution.
Area of one triangle = \(\frac{1}{2} \times 6 \times 6 \times \text{sin } 45^\circ = 12.73 \text{ cm}^2\text{.}\)
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Total area = \(8 \times 12.73 = 101.84 \text{ cm}^2\text{.}\)
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3.
In the following diagram, an octagon withside length \(s\) has been drawn, and a square has been placed around it. By finding the area of the cut off corners, or otherwise, show that: \(\text{Octagon Area} = 2(1+\sqrt{2})s^2\text{.}\)
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Solution.
Start with a square that contains the octagon: cut off four identical right isosceles triangles from the corners of a square to form the regular octagon.
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Let the octagon side be \(s\text{.}\) Each cut-off triangle is right isosceles with hypotenuse equal to the octagon side \(s\text{,}\) so each leg has length \(x = \tfrac{s}{\sqrt{2}}\text{.}\)
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The side length of the containing square is \(L = s + 2x = s + 2\left(\tfrac{s}{\sqrt{2}}\right) = s(1+\sqrt{2})\text{.}\)
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Area of the square is \(L^2 = s^2(1+\sqrt{2})^2 = s^2(3+2\sqrt{2})\text{.}\) The four removed triangles together have area \(4\times \tfrac{1}{2}x^2 = 2x^2 = 2\left(\tfrac{s^2}{2}\right) = s^2\text{.}\)
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Therefore the octagon area is \(A = L^2 - s^2 = s^2(3+2\sqrt{2}) - s^2 = s^2(2+2\sqrt{2}) = 2(1+\sqrt{2})s^2.\)
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4.
The area of a regular octagon is \(77.25 \text{ cm}^2\text{.}\) Using the formula \(\text{Area} = 2(1+\sqrt{2})s^2\text{,}\) find the side length to the nearest tenth of a centimetre.
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Solution.
Using \(A = 2(1+\sqrt{2})s^2\text{,}\) solve for \(s\text{:}\)
\begin{gather*} s = \sqrt{\dfrac{A}{2(1+\sqrt{2})}} \end{gather*}
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Substitute the numbers: \(s = \sqrt{\dfrac{77.25}{2(1+1.414)}} = \sqrt{\dfrac{77.25}{4.828}} \approx \sqrt{16} = 4 \text{ cm}.\)
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Checkpoint 2.5.43.

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Checkpoint 2.5.44.

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