Question

Quadrilateral ABCD is a trapezium of area 24.5 cm . Side AD parallel side BC , angle DAB = 90°,AD= 10 cm and BC= 4cm. If ABE is a quadrant of circle find the area of shaded region π=22\7

Answer 1

we know Area of the trapezium =
$\frac{h}{2} (a + b)$
where a and b are the parallel sides and h is the perpendicualr distance between the two parallel sides.
now
Area of the trapezium ABCD =
$\frac{AB}{2} (AD + BC) = \frac{AB }{2} (10 + 4) = 7AB$
now
$7AB = 24.5 \\ or. \: AB = \frac{24.5}{7} = 3.5$
Radius of the quadrant of the circle = AB = 3.5 cm
Area of the quadrant of the circle =
$\frac{1}{4} \pi {r}^{2} \\ = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 \\ = 9.625 \: {cm}^{2}$
 Area of the shaded region = Area of the trapezium - Area of the quadrant of the circle   = 24.5 - 9.625   = 14.875 ${cm}^{2}$