Area of a Triangle by Base and Height

The most familiar area formula in school mathematics is $\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height}.$ This lesson reviews the formula, recalls why it is true, and treats a few special cases (right, equilateral, isosceles) , so you can compute triangle areas instantly when the height is known.

Definitions

In a triangle, any one of the three sides may be designated the base. The height (or altitude) corresponding to that base is the perpendicular distance from the opposite vertex to the line containing the base.

A triangle has three different (base, height) pairs , one for each side. Whichever pair you choose, the area formula gives the same answer.

The formula and why it works

Formula. For a triangle with base $b$ and corresponding height $h$, $\text{Area} = \frac{1}{2} \times b \times h.$

Why $\tfrac{1}{2}$? Imagine the triangle inside a rectangle of width $b$ and height $h$. The diagonal of the rectangle divides it into two congruent right triangles, each with area $\tfrac{1}{2} \times b \times h$. So a right triangle satisfies the formula. For a general triangle, slide the apex along a line parallel to the base , the area does not change (this is the shearing principle). So any triangle of base $b$ and height $h$ has the same area as a right triangle with the same base and height: $\tfrac{1}{2} b h$.

Three (base, height) pairs. A triangle with sides $a, b, c$ has three corresponding heights $h_a, h_b, h_c$ such that $\text{Area} = \tfrac{1}{2} a h_a = \tfrac{1}{2} b h_b = \tfrac{1}{2} c h_c.$ So if you know any one side and the height to that side, you get the area.

Special-case formulas

These follow from the general formula by computing the relevant height.

Right triangle. If the right angle is at $C$, the two legs $a$ and $b$ play the roles of base and height: $\text{Area} = \tfrac{1}{2} \times a \times b.$ No square root or extra computation.

Equilateral triangle. All sides equal $a$. The altitude from any vertex to the opposite side has length $\tfrac{\sqrt{3}}{2} a$ (by Pythagoras in the half-triangle). So $\text{Area} = \tfrac{1}{2} \times a \times \tfrac{\sqrt{3}}{2} a = \tfrac{\sqrt{3}}{4} a^2.$

Isosceles triangle. Equal sides $b$, base $a$. The altitude from the apex bisects the base (since it's isosceles), so by Pythagoras the altitude is $\sqrt{b^2 - (a/2)^2} = \tfrac{1}{2}\sqrt{4b^2 - a^2}$. Hence $\text{Area} = \tfrac{1}{2} \times a \times \tfrac{1}{2}\sqrt{4b^2 - a^2} = \tfrac{a}{4} \sqrt{4b^2 - a^2}.$

Finding the height when only sides are known

If you know all three sides but no heights, you have two ways to find the area:

1. Use Heron's formula directly (next lesson).
2. Compute one height using Pythagoras (only for special triangles like right, equilateral, isosceles), then apply $\tfrac{1}{2} b h$.

In general, the side lengths alone are not enough to find the height without Heron's formula (or trigonometry, which you'll see later). That is why Heron's formula is so useful.

Worked examples

Example 1. A triangle has base $10$ cm and height $6$ cm. Find the area.

$\text{Area} = \tfrac{1}{2} \times 10 \times 6 = 30$ cm$^2$.

Example 2. A right triangle has legs $3$ cm and $4$ cm. Find the area.

$\text{Area} = \tfrac{1}{2} \times 3 \times 4 = 6$ cm$^2$.

Example 3. An equilateral triangle has side $6$ cm. Find the area.

$\text{Area} = \tfrac{\sqrt{3}}{4} \times 6^2 = \tfrac{\sqrt{3}}{4} \times 36 = 9\sqrt{3}$ cm$^2$ $\approx 15.59$ cm$^2$.

Example 4. An isosceles triangle has base $8$ cm and equal sides $5$ cm each. Find the area.

By the formula, $\text{Area} = \tfrac{8}{4}\sqrt{4 \cdot 25 - 64} = 2\sqrt{36} = 2 \times 6 = 12$ cm$^2$.

Example 5. A triangle has area $24$ cm$^2$ and base $8$ cm. Find the height.

$24 = \tfrac{1}{2} \times 8 \times h \Rightarrow h = 6$ cm.

Try it yourself

1. Find the area of a triangle with base $12$ cm and height $5$ cm.
2. Find the area of a right triangle with legs $6$ and $8$ cm.
3. Find the area of an equilateral triangle with side $4$ cm.
4. Find the area of an isosceles triangle with base $10$ cm and equal sides $13$ cm each.
5. A triangle has area $50$ cm$^2$ and base $10$ cm. Find the height.
6. A triangle has area $30$ cm$^2$ and height $6$ cm. Find the base.
7. Derive the formula for the area of an equilateral triangle of side $a$.
8. Show that the three (base, height) products for a triangle are all equal.
9. A right triangle has hypotenuse $10$ and one leg $6$. Find the area.
10. Find the area of an equilateral triangle of side $10$ cm.

Pitfalls / Insight

• The height must be perpendicular to the chosen base. Don't confuse a slant side with a height.
• You can choose any side as the base. The result is the same; pick the most convenient one.
• Right triangles are simplest. The two legs play the roles of base and height.

Insight. $\tfrac{1}{2} b h$ is the universal triangle-area formula. It is exact, simple, and works in every case. The challenge in problems is usually finding the height , and that is precisely where Heron's formula will help in the next lesson.