For the trapezium given below; find its area.

Draw DE and CF perpendicular to AB.

ABCD is an isosceles trapezium.

Let EF = FB = x cm

DC = EF = 18 cm

AB = AE + EF + FB

⇒ 30 = x + 18 + x

⇒ 30 = 2x + 18

⇒ 2x = 30 - 18

⇒ 2x = 12

⇒ x = $\dfrac{12}{2}$

⇒ x = 6 cm

As EDA is a right angled triangle, by using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ (6)² + height² = 12²

⇒ 36 + height² = 144

⇒ height² = 144 - 36

⇒ height² = 108 cm

⇒ height = $\sqrt{108}$ cm

⇒ height = 10.39 cm

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

= $\dfrac{1}{2}$ x (18 + 30) x 10.39

= $\dfrac{1}{2}$ x 48 x 10.39 sq. cm

= 24 x 10.39 sq. cm

= 249.41 sq. cm

Hence, the area of trapezium ABCD is 249.41 sq. cm.