Area of a Trapezium

A trapezium, also known as a trapezoid, is a quadrilateral in which a pair of sides are parallel, but the other pair of opposite sides are non-parallel. The area of a trapezium is computed with the following formula:

\[\text{Area}=\frac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}.\]

The parallel sides are called the bases of the trapezium. Let \(b_1\) and \(b_2\) be the lengths of these bases. The distance between the bases is called the height of the trapezium. Let \(h\) be this height. Then this formula becomes:

\[\text{Area}=\frac{1}{2}(b_1+b_2)h\]

Given a trapezium, let \(b_1\) and \(b_2\) be the lengths of the bases, and let \(h\) be the height. Draw a segment parallel to the bases that is halfway between the bases. This divides the trapezium into two trapeziums, each with the same height of \(\frac{1}{2}h.\)

Labeling the angles of these trapeziums:

Note the following congruences and identities due to the fact that the bases are parallel:

\[\begin{align} m \angle 4 + m \angle 5 &= 180^\circ \\ m \angle 1 + m \angle 7 &= 180^\circ \\ \angle 2 &\cong \angle 6 \\ \angle 3 &\cong \angle 8 \end{align}\]

Now rotate the top trapezoid and place it adjacent to the bottom trapezoid in the following way:

Due to the aforementioned congruences and identities, this shape is a parallelogram. The length of its base is \((b_1+b_2),\) and its height is \(\frac{1}{2}h.\) This parallelogram has the same area as the trapezoid, so the area of the trapezoid is

\[\text{Area}=\frac{1}{2}(b_1+b_2)h.\ _\square\]

Consider a trapezium \(ABCD\) in which \(AB \parallel CD\). \(AB=10\text{ cm}\), \(CD=\text{ cm}\) and they are separated by a distance of \(4\text{ cm}\). Find the area of \(ABCD\).

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We know that the area of a trapezium = \(\dfrac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}\)
Substituting the values, we get,
Area = \(\dfrac {1}{2}×(10+5)×4\)
=\(\dfrac {1}{2}×15×4\)
=\(30\text{ cm}^{2}\). \(_\square\)

 Find the area of a trapezium with parallel lines of \(9\text{ cm}\) and \(7\text{ cm}\), and a height of \(3\text{ cm}\).

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Area =\(\dfrac {1}{2}×(9 \text{ cm} + 7 \text{ cm}) × 3 \text{ cm}\) 
=\(\dfrac {1}{2} ×(16 \text{ cm})×(3 \text{ cm})\) 
=\(\dfrac {1}{2}×48 \text{ cm}^2\)   =\(24 \text{ cm}^2\). \(_\square\)