Question

In figure, ABCD is a trapezium with $AB\parallel DC$. AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn , then find the area of the shaded region of the figure.

Answer 1

Given: AB = 18 cm, DC = 32 cm, height, (h) = 14 cm and arc of radii = 7 cm

We know that,
Area of sector with an angle $\theta$
$=\frac{\theta \times \pi r^2}{360}$

Since, $AB\parallel DC$

$\therefore \angle A+\angle D={180}^{\circ }$ And $\angle B+\angle C={180}^{\circ }$
[ Interior angles on same side of parallel lines ]

$\therefore$ Area of sector with angle A and D
$=\frac{\theta \times \pi r^2}{360}$
$=\frac{{180}^{\circ }}{360}\times \frac{22}{7}\times (7{)}^2$
$=11\times 7=77c{m}^2$

Similarly, area of sector with angle B and C
$=77c{m}^2$

Now, area of trapezium
$=\frac{1}{2}\left(sumoflengthofparllelsides\right)\times h$
$=\frac{1}{2}(AB+DC)\times h$
$=\frac{1}{2}(18+32)\times 14$
$=\frac{50}{2}\times 14=350c{m}^2$

$\therefore$ Area of shaded region = (Area of trapezium) - ( Area of sector points A and D + Area of sector points B and C)
$=350-(77+77)$
$=196c{m}^2$

Hence, the required area of shaded region is $196c{m}^2$