For the trapezium given below; find its area.

Draw CE parallel to DA which meets AB at point E.

AE = DC = 20 cm

AB = AE + EB

⇒ 32 = 20 + EB

⇒ EB = 32 - 20 = 12 cm

And, DA = CE = 10 cm

For the Δ EBC,

Let the sides of the triangle be:

a = 10 cm, b = 12 cm and c = 16 cm.

The semi-perimeter s:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{10 + 12 + 16}{2}\\[1em] = \dfrac{38}{2}\\[1em] = 19$

∵ Area of triangle EBC = $\sqrt{s(s - a)(s - b)(s - c)}$

= $\sqrt{19(19 - 10)(19 - 12)(19 - 16)}$ cm²

= $\sqrt{19 \times 9 \times 7 \times 3}$ cm²

= $\sqrt{3,591}$ cm²

= 59.9 cm²

Let h be the height of Δ EBC,

Area of Δ EBC = $\dfrac{1}{2}$ x base x height

⇒ $\dfrac{1}{2}$ x 12 x height = 59.9

⇒ 6 x height = 59.9

⇒ height = $\dfrac{59.9}{6}$

⇒ height = 9.98 cm

Area of trapezium ABCD = $\dfrac{1}{2}$ x (sum of parallel sides) x height

= $\dfrac{1}{2}$ x (20 + 32) x 9.98

= $\dfrac{1}{2}$ x 52 x 9.98 sq. cm

= 26 x 9.98 sq. cm

= 259.65 sq. cm

Hence, the area of trapezium ABCD is 259.65 sq. cm.