Question

$ABCD$ is a trapezium of area $24.5$ sq. cm. In it, $AD\parallel BC,\angle DAB={90}^{\circ },AD=10cm$ and $BC=4cm$. If $ABE$ is a quadrant of a circle, find the area of the shaded region.

$\left[\text{Take}\pi =\frac{22}{7}\right]$.

Answer 1

Given:

$AD=10$ cm

$BC=4$ cm

Area of trapezium $=24.5c{m}^2$

Area of trapezium $=\frac{a+b}{2}\times height$

$24.5=\frac{AD+BC}{2}\times AB$

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

$AB=3.5$ cm

Radius $=3.5$ cm

Area of quadrant $=\frac{\text{Area of circle}}{4}$

Area of quadrant $=\frac{1}{4}\times \pi \times r^2=\frac{22}{7}\times {3.5}^2\times 0.25=9.625$ $c{m}^2$

The area of shaded region $=$ Area of trapezium $–$ Area of given quadrant

The area of shaded region $=24.5–9.625=14.875c{m}^2$