Consider the trapezium ABCD,
where AB = 6 cm, BC = 12 cm, CD = 8 cm and AD = 22 cm.
Draw a line from C parallel to AB at E such that ABCE is a parallelogram.
∴ CE = AB = 6 cm and AE = BC = 12 cm.
AD = AE + ED
⇒ ED = AD – AE
⇒ ED = 22 – 12 = 10 cm.
Area of Trapezium ABCD = Area of Parallelogram ABCE + Area of ΔCED
Draw a line through C to ED such that CF is perpendicular to ED.
Area of ΔCED:
From Heron’s formulae,
Δ CED = √[(s) (s - a) (s - b) (s - c)], where s = (a + b + c)/2 and a, b, c are side of the triangle.
In this case, a = 6; b = 8 and c = 10
⇒ s = 24/2 = 12
Area of ΔCED = √(12 × 6 × 4 × 2)
⇒ Area of ΔCED = √(576) = 24 sq. cm.
Area of ΔCED = (1/2) × ED × CF
⇒ 24 = (1/2) × 10 × CF
⇒ CF = 4.8 cm.
Area of Parallelogram ABCE = AE × CF
⇒ Area of Parallelogram ABCE = 12 × 4.8 = 57.6 sq. cm.
Area of trapezium ABCD = Area of parallelogram ABCE + Area of ΔCED
⇒ Area of trapezium ABCD = 57.6 + 24 = 81.6 sq. cm.
∴ Approximate area of the trapezium = 81.6 sq. cm.