Question

In fig., $\square$ ABCD is a trapezium of A.24.5 sq cm side ADll side BC, $\angle DAB={90}^o$ AD =10 cm and BC=4 cm, If ABE is a quadrant of a circle, find
the area of the shaded region $(\pi =\frac{22}{7})$

Answer 1

$\Rightarrow$ Area of trapezium$=24.5$

$\Rightarrow \frac{1}{2}(AD+BC)AB=24.5$

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

$\Rightarrow 7AB=24.5$

$\Rightarrow AB=\frac{24.5}{7}$

$\Rightarrow AB=3.5$cm

$\Rightarrow$ Radius of circle(Quadrant)

$=3.5$cm $=\frac{7}{2}$cm

Its area$=\frac{1}{4}\left(\pi r^2\right)=\frac{1}{4}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}$

$=\frac{77}{8}c{m}^2=9.625$

$\Rightarrow$ Area of shaded region$=24.5-9.625=14.875c{m}^2$.