How To Find The Area Of A Trapezium

Proof
Computing Area

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

Labeling the angles of these trapeziums:

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

$\begin{aligned} m \angle 4 + thou \angle five &= 180^\circ \\ g \angle 1 + 1000 \angle seven &= 180^\circ \\ \bending 2 &\cong \angle 6 \\ \angle 3 &\cong \bending eight \end{aligned}$

Now rotate the tiptop trapezoid and place it adjacent to the bottom trapezoid in the following mode:

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

$\text{Area}=\frac{one}{ii}(b_1+b_2)h.\ _\foursquare$

Consider a trapezium $ABCD$ in which $AB \parallel CD$ . $AB=x\text{ cm}$ , $CD=\text{ cm}$ and they are separated by a distance of $iv\text{ cm}$ . Find the area of $ABCD$ .

Nosotros know that the area of a trapezium = $\dfrac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}$
Substituting the values, nosotros become,
Area = $\dfrac {1}{ii}×(10+5)×four$
= $\dfrac {1}{two}×15×4$
= $30\text{ cm}^{two}$ . $_\square$

Find the surface area of a trapezium with parallel lines of $9\text{ cm}$ and $vii\text{ cm}$ , and a superlative of $3\text{ cm}$ .

Surface area = $\dfrac {1}{2}×(9 \text{ cm} + vii \text{ cm}) × 3 \text{ cm}$
= $\dfrac {1}{2} ×(xvi \text{ cm})×(3 \text{ cm})$
= $\dfrac {one}{2}×48 \text{ cm}^2$ = $24 \text{ cm}^2$ . $_\square$