Area of a Trapezoid — Formula & Step-by-Step

Use the area of a trapezoid formula

$$A = \frac{1}{2}(b_1 + b_2)h$$

Here, $b_1$ and $b_2$ are the two parallel sides, and $h$ is the perpendicular height between them. If the given side is slanted instead of perpendicular, it is not the height for this formula.

Area of a trapezoid formula

Another way to write the same formula is

$$A = \left(\frac{b_1 + b_2}{2}\right)h$$

This shows the main idea: a trapezoid acts like a rectangle whose width is the average of the two parallel sides. That is why you add the bases, divide by $2$, and then multiply by the height.

If the two parallel sides were equal, the trapezoid would become a rectangle. The formula would reduce to

$$A = \frac{1}{2}(b + b)h = bh$$

That is a quick check that the formula is reasonable.

Worked example with bases $8$ cm and $14$ cm

Suppose a trapezoid has parallel sides of $8$ cm and $14$ cm, and a perpendicular height of $5$ cm.

Start with the formula:

$$A = \frac{1}{2}(b_1 + b_2)h$$

Substitute the values:

$$A = \frac{1}{2}(8 + 14)(5)$$

Add the parallel sides:

$$A = \frac{1}{2}(22)(5)$$

Multiply and simplify:

$$A = 11 \cdot 5 = 55$$

So the area is

$$55\ \text{cm}^2$$

A fast check helps here. The average of $8$ and $14$ is $11$, so the trapezoid should match a rectangle with width $11$ cm and height $5$ cm. That also gives $55\ \text{cm}^2$.

Common mistakes when finding trapezoid area

1. Using a non-parallel side in place of one of the bases.
2. Using a slanted side as the height when it is not perpendicular.
3. Forgetting the factor of $\frac{1}{2}$.
4. Multiplying only one base by the height instead of using both parallel sides.
5. Writing the answer in plain units instead of square units.

When the area of a trapezoid is used

This formula appears in geometry class, composite-shape problems, floor plans, and land-measurement diagrams. It also shows up in coordinate geometry when a four-sided figure has one pair of parallel sides.

In applied problems, the key is identifying the correct pair of parallel sides and the true perpendicular height. If those are chosen correctly, the calculation is usually straightforward.

Try a similar problem

Try your own version with parallel sides $6$ m and $10$ m and height $4$ m. Then change only the height and solve it again. If you want one more case after that, compare what changes when the bases change but the height stays fixed.