Question

In Figure, $ABCD$ is a trapezium of area $24.5$ sq. cm. In it, $AD\parallel BC$, $\angle DAB={90}^o,AD=10$ cm and $BC=4$ cm. If $ABE$ is a quadrant of a circle, then find the area of the shaded region (upto the nearest interger). $[\text{Take}\pi =\frac{22}{7}]$

Answer 1

Given: trapezium $ABCD$, $\angle DAB={90}^{\circ }$, $AD=10$, $BC=4$ cm ,$Area=24.5sq.cm$

Area of trapezium $=$ $\frac{1}{2}$ (Sum of parallel sides) (Perpendicular)

$24.5=\frac{1}{2}\times (10+4)\times AB$

$AB=3.5$

Area of quadrant $ABE=$ $\frac{\theta }{360}\pi (AB{)}^2$

Area of quadrant $ABE=$ $\frac{90}{360}\times \frac{22}{7}(3.5{)}^2$

Area of quadrant $ABE=$ $9.625{\text{cm}}^2$

Area of shaded portion $=$ Area of trapezium $-$ Area of quadrant $ABE$

Area of shaded portion $=$ $24.5-9.625$

Area of shaded portion $=$ $14.375{\text{cm}}^2$ $=14{\text{cm}}^2$(rounding off to the nearest integer)