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How to find the area of a trapezoid

The rule: A = (b₁ + b₂)/2 × h

A trapezoid has two parallel sides - the bases $b_{1}$ and $b_{2}$ - and the height $h$ is the perpendicular distance between them. The area is the average of the bases times the height:

A = \frac{b _{1} + b _{2}}{2} \cdot h

Why the average: two copies of the trapezoid put together make a parallelogram with base $b_{1} + b_{2}$ and height $h$ , so one trapezoid is half of that. The usual trap: a worksheet figure often shows a slanted side as well. That side is not the height and takes no part in the formula - the height is the segment perpendicular to both bases.

Problem

Two worksheet tasks of the same type. Example 1. Find the area of a right trapezoid with bases $8 yd$ and $20 yd$ , height $5 yd$ and a slanted side of $13 yd$ . Example 2. Find the area of a trapezoid with a top base of $7 in$ , a bottom base of $15 in$ , a height of $4 in$ and a left side of $5 in$ .

Example 1: the bases are 8 yd and 20 yd, the height is the vertical 5 yd side. The slanted 13 yd side plays no part in the area formula.

Answer

First example:

A = 70 yd^{2}

. Second example:

A = 44 in^{2}

Checked by hand: both areas recomputed from the formula.

Step-by-step solution

Example 1: bases 8 yd and 20 yd, height 5 yd

The parallel sides are $b_{1} = 8 yd$ and $b_{2} = 20 yd$ , and the height - the side perpendicular to both - is $h = 5 yd$ . The slanted $13 yd$ side is not used.

A = \frac{b _{1} + b _{2}}{2} \cdot h = \frac{8 + 20}{2} \cdot 5

A = \frac{28}{2} \cdot 5 = 14 \cdot 5 = 70

The sides were measured in yards, so the area is in square yards:

A = 70 yd^{2}

Example 2: bases 7 in and 15 in, height 4 in

Same formula, same reasoning - and again the $5 in$ side is a decoy, not the height.

A = \frac{7 + 15}{2} \cdot 4 = \frac{22}{2} \cdot 4 = 11 \cdot 4 = 44

A = 44 in^{2}

How to check yourself

The answer must lie between the two rectangles you can draw on the bases: $b_{1} \cdot h$ and $b_{2} \cdot h$ . In the first example that is between $8 \cdot 5 = 40$ and $20 \cdot 5 = 100$ - and $70$ sits comfortably inside. If your answer falls outside that range, you have used the slanted side instead of the height.

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