Grade 10 · Core Mathematics · Lesson 8 of 12

Area of Regular Polygons: Heptagons and Octagons (Lesson 2)

A Grade 10 Core Mathematics CBE lesson from the sub-strand Area of Polygons. In it, learners determine the total area of regular octagons and heptagons given side lengths in practical contexts.

Strand Measurements and GeometrySub-strand Area of PolygonsLesson code 10CM-25-L840 minutes · 2 slides

KICD learning outcome (e)

Work out the area of regular heptagon and octagon.

By the end of this lesson, learners should be able to

Determine the total area of regular octagons and heptagons given side lengths in practical contexts.
Perform cost calculations for real-world polygonal surfaces based on area.
Calculate unknown side lengths of regular heptagons and octagons given their total surface area.

Key vocabulary

Apothem: The perpendicular distance from the centre of a regular polygon to any of its sides.; The altitude of each central triangle represents the apothem of the regular octagonal gazebo floor.
Inverse Calculation: Reversing a standard formula to solve for an unknown input from a known output.; We perform an inverse calculation to find side length from the total heptagon area.

Lesson preview

Slide 1 · Calculating Areas of Regular Heptagons and Octagons from Side LengthsModel · 18 min

Figure 1: A regular octagon is divided into eight central triangles, highlighting one isosceles triangle with its altitude, half-angle, and base measurements.

To find the area of a regular octagon or heptagon when given its side length, we decompose the polygon into congruent central triangles. By halving a central isosceles triangle, we form a right-angled triangle. We use the tangent trigonometric ratio with the half-angle and half-side to calculate the triangle's altitude, find the area of one central triangle, and multiply by the total number of sides.

Dividing a regular octagon into eight congruent triangles gives a central angle of $45^{\circ}$ for each triangle.
Halving the central angle forms a right-angled triangle where altitude is found using the tangent trigonometric ratio.
The total area equals eight times the area of one central triangle computed using base and altitude.
Tiling costs are calculated by multiplying total surface area in square metres by cost per square metre.

Worked example. Central angle for one triangle =

360^{\circ} \div 8 = 45^{\circ}

. Half-angle =

{22.5}^{\circ}

. Half-side length =

4 m \div 2 = 2 m

. Altitude

h = \frac{2}{\tan ({22.5}^{\circ})} \approx 4.8284 m

. Area of 1 central triangle =

0.5 \times 4 m \times 4.8284 m \approx 9.6568 m^{2}

. Total area of octagon =

8 \times 9.6568 m^{2} = 77.2544 m^{2} \approx 77.25 m^{2}

. Total tiling cost =

77.2544 m^{2} \times 600 = Ksh 46, 352.64

. Answer: Area = 77.25 m², Total Cost = Ksh 46,352.64.

Check question: A regular heptagonal patio has a side length of 14 m. Calculate its surface area to two decimal places.

101.75 m²
273.53 m²
712.25 m²
1424.49 m²

Show answer

712.25 m². Central angle = $360^{\circ} \div 7 \approx {51.4286}^{\circ}$ , half-angle $\approx {25.7143}^{\circ}$ . Altitude $h = 7 \div \tan ({25.7143}^{\circ}) \approx 14.5357 m$ . Area of 1 triangle = $0.5 \times 14 \times 14.5357 = 101.7499 m^{2}$ . Total heptagon area = $7 \times 101.7499 \approx 712.25 m^{2}$ .

Slide 2 · Solving Inverse and Real-World Problems Involving Regular PolygonsPractice · 22 min

Figure 2: A regular heptagon has one central triangle extracted, showing its area, base, and altitude expressed in terms of the side length.

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Teacher notes, SNE adaptations and assessment checks
Homework and the other 11 lessons in this sub-strand

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Area of Regular Polygons: Heptagons and Octagons (Lesson 2)

By the end of this lesson, learners should be able to

Key vocabulary

Lesson preview

Teach the full lesson on the projector

All 12 lessons in Area of Polygons