For the trapezium given below; find its area.

Draw CO such that CO is perpendicular to AB.

AO = DC = 8 cm

AB = AO + OB

⇒ 14 = 8 + OB

⇒ OB = 14 - 8 = 6 cm

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

In triangle BCO, by using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ (6)² + Height² = 10²

⇒ 36 + Height² = 100

⇒ Height² = 100 - 36

⇒ Height² = 64

⇒ Height = $\sqrt{64}$

⇒ Height = 8 cm

Area of trapezium ABCD = $\dfrac{1}{2}$ x (8 + 14) x 8

= $\dfrac{1}{2}$ x 22 x 8 sq. cm

= 11 x 8 sq. cm

= 88 sq. cm

Hence, the area of trapezium ABCD is 88 sq. cm.