Grade 10 · Core Mathematics · Lesson 3 of 12

Determining Area of a Triangle using Heron's Formula

A Grade 10 Core Mathematics CBE lesson from the sub-strand Area of Polygons. In it, learners investigate the relationship between perpendicular height area calculations and side-length calculations using semi-perimeter.

Strand Measurements and GeometrySub-strand Area of PolygonsLesson code 10CM-25-L340 minutes · 2 slides

KICD learning outcome (c)

Determine the area of a triangle using Heron's formula.

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

Investigate the relationship between perpendicular height area calculations and side-length calculations using semi-perimeter.
Calculate the area of a triangle given three side lengths using Heron's formula.

Key vocabulary

Semi-perimeter: Half of the total boundary length or perimeter of a closed geometric shape.; A triangle with sides $15 cm$ , $13 cm$ , and $14 cm$ has semi-perimeter $21 cm$ .
Heron's formula: A mathematical formula used to calculate the area of any triangle from its three side lengths.; Heron's formula gives area $70.99 m^{2}$ for sides $11 m$ , $13 m$ , and $16 m$ .

Lesson preview

Slide 1 · Investigating Triangular Area from Side LengthsOpening · 18 min

Figure 1: A triangle with side lengths of 13, 14, and 15 centimetres, showing a perpendicular height of 12 centimetres to the base.

When perpendicular height is difficult to measure directly on a physical land plot, side lengths offer a direct path to area. By calculating semi-perimeter $s = \frac{1}{2} (a + b + c)$ , we evaluate the radical expression $\sqrt{s (s - a) (s - b) (s - c)}$ . Constructing a perpendicular line segment shows that this square root formula produces the exact same numeric area as $\frac{1}{2} \times base \times height$ , validating its underlying geometric relationship.

The semi-perimeter $s$ represents half of the total perimeter of a triangle.
Perpendicular height $h$ enables calculating area via the standard formula $\frac{1}{2} c h$ .
The expression $\sqrt{s (s - a) (s - b) (s - c)}$ yields the identical area value without needing height.
Comparing both methods confirms that side lengths alone determine triangular area.

Worked example. Calculate area with base

c = 14 cm

and height

h = 12 cm

Area = \frac{1}{2} \times 14 \times 12 = 84 {cm}^{2}

. Find semi-perimeter:

s = \frac{15 + 13 + 14}{2} = 21 cm

. Calculate side differences:

s - a = 6 cm

s - b = 8 cm

s - c = 7 cm

. Evaluate radical:

\sqrt{21 \times 6 \times 8 \times 7} = \sqrt{7056} = 84 {cm}^{2}

. Answer: Both methods give an area of $84 {cm}^{2}$ ..

Check question: A triangle has side lengths $a = 10 cm$ , $b = 17 cm$ , and $c = 21 cm$ . Calculate the semi-perimeter $s$ and evaluate $\sqrt{s (s - a) (s - b) (s - c)}$ to find its area.

$s = 24 cm$ , $Area = 84 {cm}^{2}$
$s = 24 cm$ , $Area = 7056 {cm}^{2}$
$s = 48 cm$ , $Area \approx 1235.59 {cm}^{2}$
$s = 24 cm$ , $Area = 105 {cm}^{2}$

Show answer

$s = 24 cm$ , $Area = 84 {cm}^{2}$ . First, calculate the semi-perimeter $s$ by adding the side lengths and dividing by $2$ : $s = \frac{10 + 17 + 21}{2} = \frac{48}{2} = 24 cm$ . Next, apply Heron's formula: $Area = \sqrt{24 (24 - 10) (24 - 17) (24 - 21)}$ $Area = \sqrt{24 \times 14 \times 7 \times 3}$ $Area = \sqrt{7056} = 84 {cm}^{2}$ .

Slide 2 · Applying Heron's Formula to Solve Triangular Area ProblemsModel · 22 min

Figure 2: A triangle with side lengths of 11, 13, and 16 metres, alongside the initial calculation steps for using Heron's formula.

Teach the full lesson on the projector

This lesson, and every other Grade 1–10 CBE lesson, comes ready to project with the EAP ClassKit. It runs without electricity or internet.

Both slides with diagrams and class activities
Teacher notes, SNE adaptations and assessment checks
Homework and the other 11 lessons in this sub-strand

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