The two parallel sides of a trapezium are 58 m and 42 m long. The other two sides are equal, each being 17 m. Find its area.

ABCD is a trapezium.

AB = 42 m

CD = 58 m

AD = BC = 17 m

From A and B drop perpendiculars AE and BF respectively to DC.

From figure,

EF = AB = 42 m

Since, AD = BC (given) and AE = BF (perpendicular between same parallels)

Thus,

DE = FC = x (let)

From figure,

⇒ DE + FC + EF = DC

⇒ x + x + 42 = 58

⇒ 2x = 58 - 42

⇒ 2x = 16

⇒ x = 8 meters.

In △ AED,

Using pythagoras theorem,

⇒ AD² = AE² + ED²

⇒ 17² = AE² + 8²

⇒ AE² = 17² - 8²

⇒ AE² = 289 - 64

⇒ AE² = 225

⇒ AE = $\sqrt{225}$ = 15 m.

Height = AE = 15 m.

By formula,

Area of trapezium = $\dfrac{1}{2} \times \text{ (sum of // sides)} \times \text{ (distance between them)}$

$\text{Area of trapezium ABCD } = \dfrac{1}{2} \times (AB + CD) \times AE \\[1em] = \dfrac{1}{2} \times (42 + 58) \times 15 \\[1em] = \dfrac{1}{2} \times 100 \times 15 \\[1em] = 750 \text{ m}^2.$

Hence, area = 750 m².