For the trapezium given below; find its area.

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

Since, AB || DC thus AE || DC and DA || CE.

Since, opposite sides are parallel, thus AECD is a parallelogram.

Opposite sides of a parallelogram are equal, thus CE = AD = 10 cm and AE = DC = 12 cm.

In isosceles triangle EBC,

Draw CF ⊥ EB.

In an isosceles triangle, the perpendicular from the common vertex to the base, bisects it.

Thus, EF = $\dfrac{EB}{2} = \dfrac{8}{2}$ = 4 cm.

In right triangle CEF,

By pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

EC² = CF² + EF²

10² = CF² + 4²

100 = CF² + 16

CF² = 100 - 16

CF² = 84

CF = $\sqrt{84}$ = 9.16 cm

By formula,

Area of trapezium = $\dfrac{1}{2}$ x sum of parallel sides x distance between them

Area of trapezium ABCD = $\dfrac{1}{2} \times (AB + DC) \times CF$

$= \dfrac{1}{2} \times (20 + 12) \times 9.16 \\[1em] = \dfrac{1}{2} \times 32 \times 9.16 \\[1em] = 146.56 \text{ cm}^2.$

Hence, area of trapezium ABCD = 146.56 cm².