Question

In Figure 3, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and the distance between AB and DC is 14 cm. If arcs of equal radii 7 cm have been drawn, with centres A,B, C  and D, then find the area of the shaded region.

Answer 1

Here,
Radii of the arcs with centres A, B, C and D = 7 cm
And,
AB = 18 cm
DC = 32 cm 
Distance between AB and DC = 14 cm
Area of trapezium ABCD = \[\frac{1}{2}\left( 18 + 32 \right)14\]

                                         = 350 cm²

Now,
Area of unshaded region = Area of sector with central ∠∠A + Area of sector with central ∠∠B + Area of sector with central ∠∠C + Area of sector with central ∠∠D

\[= \frac{\angle A}{360^o} \times \pi \times \left( 7 \right)^2 + \frac{\angle B}{360^o} \times \pi \times \left( 7 \right)^2 + \frac{\angle C}{360^o} \times \pi \times \left( 7 \right)^2 + \frac{\angle D}{360^o} \times \pi \times \left( 7 \right)^2 \]
\[ = \pi \times \left( 7 \right)^2 \left( \frac{\angle A}{360^o} + \frac{\angle B}{360^o} + \frac{\angle C}{360^o} + \frac{\angle D}{360^o} \right)\]
\[ = \frac{22}{7} \times \left( 7 \right)^2 \left( \frac{\angle A + \angle B + \angle C + \angle D}{360^o} \right)\]
\[ = 154\left( \frac{360^o}{360^O} \right) \left( \text{Angle sum property of quadrilateral} \right)\]
\[ = 154 {cm}^2\]

∴ Area of the shaded region = Area of trapezium ABCD − Area of unshaded region
                                                    = 350 − 154

                                                    = 196 cm²