Question

Two parallel sides of a trapezium are 58 cm and 42 cm. The other two sides are of equal length which is 17cm. Find the area of the trapezium.

• A. $1500c{m}^2$
• B. $750cm$
• C. $325c{m}^2$
• D. $750c{m}^2$

Answer 1

The correct option is D

$750c{m}^2$

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; AE = DC = 42 cm.

Now, in triangle EBC, we have CE = BC = 17 cm.
Also, CF ⟂ AB
Therefore, EF = ¹/₂ × EB = 8 cm.
We have CE = 17 cm, EF = 8 cm.
So, CF will be
$CF=\sqrt{C{E}^2-E{F}^2}$
= √(17² - 8²)
= √225 = 15 cm.

Thus, the distance between the parallel sides is 15 cm.
Area of trapezium ABCD
= ¹/₂ × (sum of parallel sides) × (distance between them)
= ¹/₂ × (58 + 42) × 15 cm
$=750c{m}^2$