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A trapezium $$ABCD$$ has side $$AD$$ parallel to $$BC, \angle BAD = 90^\circ, BC = 3$$ cm and $$AD= 8$$ cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is
Correct Answer: 66
CD = $$\sqrt{\ y^2+25}$$
$$11+y+\sqrt{y^2+25}=36$$
$$\sqrt{y^2+25}=25-y$$
$$y^2+25=25^2+y^2-50y$$
2y = 24
y = 12
Area of trapezium = $$3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$$
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