Question

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

Answer 1

As we know that area of trapezium $=\frac{1}{2}\times \left(sumofparallelsides\right)\times distancetheparallelsides$
Hence according to the figure area $=\frac{1}{2}\times (10cm+4cm)\times r=24.5c{m}^2$
$\Rightarrow \frac{1}{2}\times 14cm\times r=24.5c{m}^2$
$\Rightarrow r=\frac{24.5c{m}^2\times 2}{14cm}=3.5cm$
Now, area of quadrant $=\frac{areaofcircle}{4}$
Hence area of quadrant ABE $=\frac{areacirclewhoseradiusis3.5cm}{4}$
$=\frac{\pi \times 3.5cm\times 3.5cm}{4}=\frac{\frac{22}{7}\times 3.5cm\times 3.5cm}{4}=9.625c{m}^2$
Hence area of region DEBC = area of trapezium ABCD - area of quadrant ABE
$24.5c{m}^2-9.625c{m}^2=14.875c{m}^2$