Question

In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7cm; AD = BC = 5cm and the distance between AB and DC is 4cm. Find the length of DC and hence, find the area of trap. ABCD.

Answer 1

Consider △ ALD

Based on the Pythagoras theorem

AL² + DL² = AD²

By substituting the values

4² + DL² = 5²

So we get

DL² = 5² – 4²

DL² = 25 – 16

By subtraction

DL² = 9

By taking square root

DL = √ 9

So we get

DL = 3 cm

Consider △ BMC

Based on the Pythagoras theorem

MC² + MB² = CB²

By substituting the values

MC² + 4² = 5²

So we get

MC² = 5² – 4²

MC² = 25 – 16

By subtraction

MC² = 9

By taking square root

MC = √ 9

So we get

MC = 3 cm

From the figure we know that LM = AB = 7cm

So we know that CD = DL + LM + MC

By substituting the values

CD = 3 + 7 + 3

By addition

CD = 13cm

Area of Trapezium ABCD = ½ (sum of parallel sides × distance between them)

So we get

Area of Trapezium ABCD = ½ × (CD + AB) × AL

By substituting the values

Area of Trapezium ABCD = ½ × (13 + 7) × 4

On further calculation

Area of Trapezium ABCD = 20 × 2

By multiplication

Area of Trapezium ABCD = 40 cm²

Therefore, length of DC = 13cm and area of trapezium ABCD = 40 cm².