Question

In Fig. ABCD is a trapezium of area $24.5c{m}^2$. In it, $AD\parallel BC$, $\angle DAB={90}^{\circ }$, AD = 10 cm and BC = 4 cm. If ABE is a quadrant of a circle, find the area of the shaded region. (Take $\pi =\frac{22}{7}$)

Answer 1

Sol:
Area of the trapezium = $\frac{h(a+b)}{2}$ where a and b are the parallel sides and h is the perpendicualr distance between the parallel sides.
Area of the trapezium ABCD = $\frac{AB(AD+BC)}{2}$

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

$49=AB\times 14$

$AB=3.5cm$

Radius of the quadrant of the circle = AB = 3.5 cm
Area of the quadrant of the circle =

$\frac{\pi r^2}{4}$

$=\frac{22}{7}\times \frac{3.5\times 3.5}{4}=9.625c{m}^2$

Area of the shaded region = Area of the trapezium - Area of the quadrant of the circle

= 24.5 - 9.625

= 14.875 $c{m}^2$