From the given figure, find the area of trapezium ABCD.

From figure,

AB || DC

Distance between parallel sides, AE = BC = 4 cm

In right angled △ AED,

Hypotenuse² = Perpendicular² + Base²

⇒ AD² = AE² + ED²

⇒ 5² = 4² + ED²

⇒ 25 = 16 + ED²

⇒ ED² = 25 - 16

⇒ ED² = 9

⇒ ED = $\sqrt{9}$

⇒ ED = 3 cm.

∴ DC = EC - ED = 5 - 3 = 2 cm

As we know,

Area of trapezium = $\dfrac{1}{2} \times (\text{sum of parallel sides} \times \text{distance between them})$

$= \dfrac{1}{2} \times (\text{AB + DC}) \times \text{BC} \\[1em] = \dfrac{1}{2} \times (5 + 2) \times 4 \\[1em] = \dfrac{1}{2} \times 7 \times 4 \\[1em] = 7 \times 2 \\[1em] = 14 \text{ cm}^2.$

Hence, the area of trapezium ABCD is 14 cm².