Area of triangles

A triangle looks awkwardly slanted, so it's natural to wonder how to find its area. The answer is surprisingly simple , and surprisingly elegant.

Concept

For any triangle, choose one side to be the base and call its length $b$ . The height $h$ is the perpendicular distance from the opposite vertex to the base (or to the base extended). Then:

$\text{Area of triangle} = \frac{1}{2} \times b \times h$

The " $\frac{1}{2}$ " is the puzzling part. Where does it come from?

Proof by doubling. Take any triangle. Make an identical copy by rotating it $180°$ . Place the copy next to the original so that two sides match. The two triangles now form a parallelogram with base $b$ and height $h$ . The parallelogram's area is $b \times h$ . So each triangle is half of that:

$\text{Triangle area} = \frac{1}{2} b h$

This proof works for any triangle , right, acute, or obtuse.

A few special cases

Right triangle with legs $a$ and $b$ (the two sides meeting at the right angle): area $= \frac{1}{2} a b$ . The legs serve as base and height.
Equilateral triangle with side $s$ has height $h = \frac{\sqrt{3}}{2}s$ , so area $= \frac{\sqrt{3}}{4} s^2$ . (You will derive this later; for now the formula is enough.)
Isosceles triangle has two equal sides. If the base is $b$ and the equal sides are $a$ each, the height splits the base in half and you can compute $h$ using the rule of right triangles.

Choice of base

Any of the three sides can be the "base". The formula gives the same area no matter which side you choose , provided you use the matching height (the perpendicular from the opposite vertex to that side). Sometimes choosing one base makes the computation much easier than another.

Triangles that look different but have the same area

Two triangles with the same base length and the same height have the same area, even if they look very different. This is the secret behind many area puzzles.

Worked examples

Example 1. A triangle has base $12$ cm and height $5$ cm. Find its area.

Area $= \frac{1}{2} \times 12 \times 5 = 30$ cm².

Example 2. A right triangle has legs $7$ cm and $24$ cm. Find its area.

Area $= \frac{1}{2} \times 7 \times 24 = 84$ cm².

Example 3. A triangle has area $36$ cm² and base $9$ cm. Find its height.

$\frac{1}{2} \times 9 \times h = 36 \Rightarrow h = \frac{72}{9} = 8$ cm.

Example 4. A triangular field has base $40$ m and height $30$ m. Find its area in m².

Area $= \frac{1}{2} \times 40 \times 30 = 600$ m².

Example 5. A roof is shaped like a triangle with base $6$ m and height $4$ m. If $1$ m² of paint costs $\rupee 60$ , what is the total paint cost?

Area $= \frac{1}{2} \times 6 \times 4 = 12$ m².
Cost $= 12 \times 60 = \rupee 720$ .

Example 6. A rectangle is cut along its diagonal to form two triangles. The rectangle is $10$ cm by $7$ cm. What is the area of each triangle?

Rectangle area $= 70$ cm². Each triangle is half $= 35$ cm².
(Confirm with formula: base $10$ , height $7$ , area $\frac{1}{2} \times 70 = 35$ cm².)

Try it yourself

Find the area of a triangle with base $8$ cm and height $9$ cm.
A right triangle has legs $5$ cm and $12$ cm. Find its area.
A triangle has area $48$ cm² and height $6$ cm. Find its base.
The diagonal of a $14$ cm by $8$ cm rectangle splits it into two triangles. Find the area of each.
A triangular sail has base $4$ m and height $3$ m. Find its area.
A triangle has sides $5$ , $5$ , $6$ (isosceles). Find its height to the base of length $6$ , then find its area.
Two triangles share the same base ( $10$ cm) but one has height $4$ cm and the other $8$ cm. Compare their areas.
Investigate: a triangle has vertices at $(0, 0)$ , $(6, 0)$ and $(0, 4)$ on a grid. Sketch it, identify the base and height, and find its area.

Activity

Cut-and-paste proof. Cut out two identical triangles from coloured paper. Lay them down so they fit together to form a parallelogram. Measure the base and height of the parallelogram, then its area. Now compare with twice the area you compute for one triangle using $\frac{1}{2} b h$ . They should match exactly.