Area and Perimeter of a Trapezoid

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Each of the two parallel sides of a trapezoid is a base. The distance between the bases (measured perpendicular to each) is the height. To find the area of a trapezoid, we multiply the average length of the two bases by the height. In symbols, if the lengths of the bases are \(a\) and \(b\) and the height is \(h\) (see diagram), then the area, \(A\), of the trapezoid is \(A=\frac{(a+b)}{2}h\), which we can also write as \(A=\frac{1}{2}(a+b)h\).

For example, the trapezoid in the diagram with bases of length 5 cm and 9 cm and with height 3 cm has an area of \(A=\frac{5+9}{2}\cdot3=\frac{14}{2}\cdot3=7\cdot3=21\text{ square centimeters}\), which we may also write as \(21\text{ cm}^2\).

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The area formula for a trapezoid works because it comes from the formula for the area of a parallelogram. The trapezoid below (with solid sides) has bases of length \(a\) and \(b\) and height \(h\). Suppose we make a copy of it, rotate it halfway around, and place it adjoining the original trapezoid so that legs (non-parallel sides) of the same length coincide (the shaded trapezoid with dashed sides). Together these figures form a parallelogram with a base of length \(a+b\) and height \(h\). By the standard formula, the area of this parallelogram is \(\text{area}=\text{base}×\text{height}=(a+b)h\). The area of the original trapezoid is half of this, namely \(\frac{1}{2}(a+b)h\), or, equivalently, \(\frac{(a+b)}{2}h\). This same procedure works for every trapezoid.

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We measure the area of a trapezoid, like all areas, in square units–square feet, square inches, square meters, etc.—which are sometimes written \(\text{ft}^2\), \(\text{in}^2\), \(\text{m}^2\), etc. Usually, we use the square of the unit used to measure the bases and height of the trapezoid. For instance, if we measure the bases and height in centimeters, we usually give the area in square centimeters.

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The perimeter of a figure is the distance around it. We find the perimeter of a trapezoid by adding up the lengths of its four sides.

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If we do not know the lengths of all four sides of a trapezoid, sometimes we have enough other information to find the lengths of the missing sides using the Pythagorean theorem. For example, in the trapezoid in the diagram, the bases are 2 cm and 9 cm, the height is 4 cm, and the longer base sticks out past the shorter base by 3 cm on the left and 4 cm on the right. This makes sides \(c\) and \(d\) hypotenuses of right triangles with sides whose lengths we know. By the Pythagorean theorem, \(c^2=3^2+4^2=9+16=25\), so \(c=\sqrt{25}=5\text{ cm}\). Similarly, \(d^2=4^2+4^2=16+16=32\), so \(d=\sqrt{32}=\sqrt{16\cdot2}=\sqrt{16}\cdot \sqrt{2}=4\sqrt{2}≈5.7\text{ cm}\). Now we can find the perimeter, \(P\), of the trapezoid by adding up the four sides: \(P=9+5+2+4sqrt{2}=16+4\sqrt{2}≈16+5.7=21.7\text{ cm}\). This is one example of finding the perimeter of a trapezoid using the Pythagorean theorem.

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If we do not know the height of a trapezoid, sometimes we have enough other information to find the height using the Pythagorean theorem. For example, in the trapezoid in the diagram, the bases are 2 cm and 9 cm, the longer base sticks out 3 cm past the shorter base on the left, and the left leg (the non-parallel side) is 5 cm. This makes the dashed line a side of a right triangle whose hypotenuse and other side we know. By the Pythagorean theorem, \(h^2+3^2=5^2\) or \(h^2+9=25\). So, \(h^2=16\) and \(h=\sqrt{16}=4\text{ cm}\), which is the height of the trapezoid. Now we can apply the standard formula to find the area of this trapezoid: \(A=\frac{(a+b)}{2}h=\frac{(2+9)}{2}\times4=\frac{11}{2}\times4=\frac{11}{1}⋅2=22\text{ cm}^2\). This is an example of finding the area of a trapezoid without the height.