Question

ABCD is a trapezium with AB and CD parallel. If AB$=6$, BC$=5$, CD$=3$, DA$=4$, A$={90}^o$ the area of ABCD is?

• A. $27$
• B. $12$
• C. $18$
• D. $15$

Answer 1

The correct option is C $18$

as $\angle A={90}^{\circ }$ Hence $DA$ is the height of the trapezium as $AB$ and $CD$ are parallel.

Area of the trapezium = $\frac{6+3}{2}\times 4$

$⟹18sq.units$