Area of Trapezoid

A trapezoid has bases of lengths \(10\,\text{cm}\) and \(15\,\text{cm}\). The perpendicular distance between the bases is \(8\,\text{cm}\). Find the area of the trapezoid.

Solution

To solve this problem, we simply need to substitute the values of the lengths of the bases and the height in the area of trapezoid formula.

\[\begin{align}\text{Area} & = \frac{1}{2} h (a + b) \\ \\& = \frac{1}{2} \cdot 8 (10 + 15) \\ \\& = 4 \cdot 25 \\ \\& = 100\,\text{cm}^2\end{align}\] The area of the trapezoid is \( 100\,\text{cm}^2 \).

Find the area of the following trapezoid.

Fig. 4. Trapezoid on the coordinate plane.

Solution

In this case, to be able to find the area of the above trapezoid, we need to find the length of the bases and the height of the trapezoid.

These values are not given, but we can use the coordinate plane to work them out.

We need to calculate the distance between each of the points, how can we do this?

The distance between the points \(B(6, 2)\) and \(C(9, 2)\) can be calculated by finding the absolute value of the difference between their x-coordinates, using \(|x_2 - x_1|\). The same applies for the distance between the points \(A(2, 7)\) and \(D(10, 7)\).

The distance between the points \(B(6, 2)\) and \(E(6, 7)\) can be calculated by finding the absolute value of the difference between their y-coordinates, using \(|y_2 - y_1|\).

\[\begin{align}a &= \overline{BC} = |x_2 - x_1| = |9 - 6| = 3 \\ \\b &= \overline{AD} = |x_2 - x_1| = |10 - 2| = 8 \\ \\h &= \overline{BE} = |y_2 - y_1| = |7 - 2| = 5 \end{align}\]Now that we have all the values that we need, we can substitute them in the area of trapezoid formula.

\[\begin{align}\text{Area} & = \frac{1}{2} \cdot 5 (3 + 8) \\ \\& = \frac{5}{2} \cdot (11) \\ \\& = \frac{55}{2} \\ \\& = 27.5\,\text{units}^2\end{align}\] The area of the trapezoid is \( 27.5\,\text{units}^2 \).

A trapezoid with an area of \(35\,\text{m}^2\) has bases of lengths, \(3\,\text{m}\) and \(4\,\text{m}\). Find the distance between the parallel sides.

Solution

The distance between the parallel sides is the height of the trapezoid. So, let's substitute the values that we have in the area of trapezoid formula, and then solve for \(h\).

\[\begin{align}\text{Area} & = \frac{1}{2} h (a + b) \\ \\35 & = \frac{1}{2} \cdot h (3 + 4) \\ \\35 & = \frac{7 \cdot h}{2} \\ \\h & = \frac{35 \cdot 2}{7} \\ \\& = \frac{70}{7} \\ \\& = 10\,\text{m}\end{align}\]

The height of the trapezoid is \(10\, \text{m}\).

Find the area of the following trapezoid.

Fig. 5. Trapezoid with no height example.

Fig. 6. Height of a trapezoid using Pythagoras.

Notice that we have a right triangle at each side. The bases of each triangle was calculated by finding the difference between \(18\) and \(10\), and then dividing the result by \(2\).

\[18 - 10 = \frac{8}{2} = 4\, \text{m}\]

Now we can calculate the height using the Pythagoras Theorem which states that, in a right triangle, the square of the hypothenuse equals the sum of the other two sides squared.

\[\begin{align}c^2 & = a^2 + b^2 \\ \\6^2 & = 4^2 + h^2 \\ \\36 & = 16 + h^2 \\ \\h^2 & = 36 - 16 \\ \\h & = \sqrt{20}\\ \\& = 4.47\, \text{m}\end{align}\]Now that we know the length of the height, we can calculate the area of the trapezoid.

\[\begin{align}\text{Area} & = \frac{1}{2} \cdot 4.47 (10 + 18) \\ \\& = \frac{1}{2} \cdot 4.47 \cdot 28 \\ \\& = 62.6\, \text{m}^2\end{align}\] The area of the trapezoid is \(62.6\, \text{m}^2\).

The length of the diagonals of a trapezoid \(d_1\) and \(d_2\) is \(6.4\, \text{m}\), and the angle \(\alpha\) between them measures \(77.3^{\circ}\). Find the area of the trapezoid.

\[\text{Area} = \frac{1}{2} \cdot 6.4 \cdot 6.4 \cdot \sin(77.3^{\circ}) = 19.98\, \text{m}^2\]

The area of the trapezoid is \(19.98\, \text{m}^2\).