For the trapezium given below; find its area.

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

DC = AE = 12 cm

AB = AE + EB

⇒ 20 = 12 + EB

⇒ EB = 20 - 12 = 8 cm

And, AD = CE = 10 cm

Area of isosceles triangle EBC =

$= \dfrac{1}{4} \times b \times \sqrt{4s^2 - b^2}\\[1em] = \dfrac{1}{4} \times 8 \times \sqrt{4 \times 10^2 - 8^2}\\[1em] = 2 \times \sqrt{4 \times 100 - 64}\\[1em] = 2 \times \sqrt{400 - 64}\\[1em] = 2 \times \sqrt{394}\\[1em] = 2 \times 19.85\\[1em] = 39.7 \text{ cm}^2$

Also, area of triangle EBC = $\dfrac{1}{2}$ x base x height

$⇒ \dfrac{1}{2} \times 8 \times \text{height} = 39.7\\[1em] ⇒ 2 \times \text{height} = 39.7\\[1em] ⇒ \text{height} = \dfrac{39.7}{2}\\[1em] ⇒ \text{height} = 19.85 \text{ cm}$

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

= $\dfrac{1}{2}$ x (12 + 20) x 9.15 sq. cm

= $\dfrac{1}{2}$ x 32 x 9.15 sq. cm

= 16 x 9.15 sq. cm

= 146.64 sq. cm

Hence, the area of trapezium is 146.64 sq. cm.