In a trapezium $$ABCD$$, $$AB$$ is parallel to $$DC$$, $$BC$$ is perpendicular to $$DC$$ and $$\angle BAD=45^{0}$$. If $$DC$$ = 5cm, $$BC$$ = 4 cm,the area of the trapezium in sq cm is
Correct Answer: 28
Given, BC = DE = 4
CD = BE = 5
In triangle ADE, EAD=45^{0}$$
$$\tan\ 45\ =\ \frac{DE}{AE}$$ => AE = 4
Area of trapezium = Area of rectangle BCDE + Area of triangle AED
= 20 + 8 = 28
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