Question

Find area of a trapezium whose parallel sides measure 58 cm and 42 cm, and the two equal sides measure 17 cm.

• A. 750

Answer 1

The correct option is A 750

Let ABCD be the given trapezium in which AB = 58 cm, DC = 42 cm, BC = 17 cm and AD = 17 cm.
Through C, draw CE parallel to AD, meeting AB at E.

Also, draw CF ⟘ AB.

Now, EB = (AB - AE) = (AB - DC)
= (58 - 42) cm = 16 cm

CE = AD = 17 cm and AE = DC = 42 cm.

Now, in triangle EBC, we have CE = BC = 17 cm.

Also, CF ⟂ AB

$\Rightarrow$ EF = ¹/₂ × EB = 8 cm

$\Rightarrow$ CE = 17 cm and EF = 8 cm

$\Rightarrow CF=\sqrt{C{E}^2-E{F}^2}=\sqrt{{17}^2-8^2}=\sqrt{225}=15cm$

Thus, the distance between the parallel sides is 15 cm.

Area of trapezium ABCD

= ¹/₂ × (sum of parallel sides) × (distance between them)

$=\frac{1}{2}\times (58+42)\times 15$

$=750c{m}^2$