Area of a Triangle — Formula and Examples

Grade: 7-8 | Topic: Geometry

What You Will Learn

After this lesson you will be able to calculate the area of any triangle when you know its base and height. You will also learn how to identify the correct height for a given base, handle right triangles as a special case, and use Heron's formula when only the three side lengths are available.

Theory

The standard area formula

Every triangle can be thought of as exactly half of a parallelogram. If you duplicate a triangle and flip the copy, the two pieces fit together to form a parallelogram with the same base and height. Because the parallelogram's area is $b \times h$, the triangle's area is half of that:

$A = \frac{1}{2} \times b \times h$

where:

• $b$ is the length of the base (any side you choose),
• $h$ is the perpendicular height — the shortest distance from the base to the opposite vertex, measured at a 90-degree angle to the base.

For example, a triangle with base 10 cm and height 6 cm has area:

$A = \frac{1}{2} \times 10 \times 6 = 30 \text{ cm}^2$

Choosing the base and height

You can pick any side of the triangle as the base. The height then changes accordingly — it is always the perpendicular distance from the chosen base to the opposite vertex. No matter which side you choose as the base, the area comes out the same.

For right triangles, the two legs are already perpendicular to each other, so one leg serves as the base and the other as the height:

$A_{\text{right}} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2$

Heron's formula — when you know all three sides

Sometimes a problem gives you all three side lengths ($a$, $b$, $c$) but no height. In that case, use Heron's formula. First calculate the semi-perimeter:

$s = \frac{a + b + c}{2}$

Then the area is:

$A = \sqrt{s(s - a)(s - b)(s - c)}$

This formula works for any triangle and avoids the need to calculate the height separately.

Worked Examples

Example 1: Basic triangle with given base and height (easy)

Problem: Find the area of a triangle with a base of 14 cm and a height of 9 cm.

Step 1: Write the formula. $A = \frac{1}{2} \times b \times h$

Step 2: Substitute the values. $A = \frac{1}{2} \times 14 \times 9 = \frac{126}{2} = 63 \text{ cm}^2$

Answer: 63 cm$^2$

Example 2: Right triangle using the two legs (easy)

Problem: A right triangle has legs of 5 m and 12 m. Find its area.

Step 1: In a right triangle the two legs are perpendicular, so use them as base and height. $A = \frac{1}{2} \times 5 \times 12$

Step 2: Calculate. $A = \frac{1}{2} \times 60 = 30 \text{ m}^2$

Answer: 30 m$^2$

Example 3: Finding the height when the area is known (medium)

Problem: A triangle has an area of 48 cm$^2$ and a base of 16 cm. What is its height?

Step 1: Start from the area formula and solve for $h$. $A = \frac{1}{2} \times b \times h \implies h = \frac{2A}{b}$

Step 2: Substitute. $h = \frac{2 \times 48}{16} = \frac{96}{16} = 6 \text{ cm}$

Answer: 6 cm

Example 4: Using Heron's formula (medium)

Problem: A triangle has sides of 7 cm, 8 cm, and 9 cm. Find its area.

Step 1: Calculate the semi-perimeter. $s = \frac{7 + 8 + 9}{2} = \frac{24}{2} = 12$

Step 2: Apply Heron's formula. $A = \sqrt{12(12 - 7)(12 - 8)(12 - 9)} = \sqrt{12 \times 5 \times 4 \times 3}$

Step 3: Simplify inside the square root. $A = \sqrt{720} = \sqrt{144 \times 5} = 12\sqrt{5} \approx 26.83 \text{ cm}^2$

Answer: $12\sqrt{5} \approx$ 26.83 cm$^2$

Example 5: Real-world application — triangular garden (challenging)

Problem: A triangular garden has a base along the sidewalk of 20 m. The perpendicular distance from the sidewalk to the back corner is 8.5 m. How many square metres of sod are needed to cover the entire garden?

Step 1: Identify the dimensions. $b = 20$ m, $h = 8.5$ m.

Step 2: Calculate the area. $A = \frac{1}{2} \times 20 \times 8.5 = \frac{1}{2} \times 170 = 85 \text{ m}^2$

Answer: 85 m$^2$ of sod is needed.

Common Mistakes

Mistake 1: Using a slant side instead of the perpendicular height

❌ A triangle with base 10 and a slant side of 8: $A = \frac{1}{2} \times 10 \times 8 = 40$

✅ The height must be perpendicular to the base. If the actual perpendicular height is 6: $A = \frac{1}{2} \times 10 \times 6 = 30$

Why this matters: The height in the area formula must form a 90-degree angle with the base. A slant side is always longer than the perpendicular height (unless the triangle is a right triangle and the slant side is a leg). Using it inflates the area.

Mistake 2: Forgetting the one-half factor

❌ $A = b \times h = 14 \times 9 = 126$

✅ $A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 14 \times 9 = 63$

Why this matters: Without the $\frac{1}{2}$, you are calculating the area of the full parallelogram, not the triangle. Your answer will be exactly double the correct area every time.

Mistake 3: Misidentifying the base and height in obtuse triangles

❌ In an obtuse triangle, the height drawn from the obtuse-angled vertex may land outside the triangle. Students sometimes ignore this and measure incorrectly.

✅ Extend the base line if needed. The perpendicular height drops from the opposite vertex to the base line (or its extension), forming a 90-degree angle.

Why this matters: Obtuse triangles can be tricky because the foot of the height falls outside the triangle's boundary. The formula still works — just extend the base to meet the height.

Practice Problems

Try these on your own before checking the answers:

1. A triangle has a base of 18 cm and a height of 7 cm. Find its area.
2. A right triangle has legs of 9 m and 40 m. Find its area.
3. The area of a triangle is 60 cm$^2$ and its base is 15 cm. Find the height.
4. A triangle has sides 5 cm, 12 cm, and 13 cm. Use Heron's formula to find the area. (Hint: check if it is a right triangle.)
5. A triangular flag has a base of 1.2 m and a height of 0.8 m. How much fabric is needed to make the flag?

Click to see answers

1. $A = \frac{1}{2} \times 18 \times 7 = 63$ cm$^2$.
2. $A = \frac{1}{2} \times 9 \times 40 = 180$ m$^2$.
3. $h = \frac{2 \times 60}{15} = 8$ cm.
4. Semi-perimeter: $s = \frac{5 + 12 + 13}{2} = 15$. $A = \sqrt{15 \times 10 \times 3 \times 2} = \sqrt{900} = 30$ cm$^2$. (This is also a right triangle with legs 5 and 12, so $A = \frac{1}{2} \times 5 \times 12 = 30$ cm$^2$.)
5. $A = \frac{1}{2} \times 1.2 \times 0.8 = 0.48$ m$^2$.

Summary

• The area of any triangle is $A = \frac{1}{2} \times b \times h$, where $h$ is the perpendicular height to the chosen base.
• For right triangles, the two legs serve as base and height directly.
• When only the three side lengths are known, use Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$.
• Always check that the height is perpendicular to the base — not a slant side.

Related Topics

• Area and Perimeter — Formulas, Examples, and Practice
• Area and Circumference of a Circle — Formulas and Examples
• How to Find the Area of a Trapezoid

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