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In the Given Figure, Abcd is a Trapezium with Ab || Dc, Ab = 18 Cm, Dc = 32 Cm and the Distance Between Ab and Ac is 14 Cm - Mathematics

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Question

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

Solution

ABCD is a trapezium with AB || DC, AB = 18 cm, CD = 32 cm and the distance between AB and AC is 14 cm.

Radii of the arcs, r = 7 cm

Now,

Area of the shaded region

= Area of trapezium ABCD − (Area of the sector of the circle with centre A + Area of the sector of the circle with centre B + Area of the sector of the circle with centre C + Area of the sector of the circle with centre D)

Area of trapezium ABCD

=\[\frac{1}{2}\]x × Sum of the parallel sides × Distance between the parallel sides
=\[\frac{1}{2}\] × (18 + 32) × 14

= 350 cm2

Also,

Area of the sector of the circle with centre A + Area of the sector of the circle with centre B + Area of the sector of the circle with centre C + Area of the sector of the circle with centre D

\[= \frac{\angle A}{360^o} \times \pi r^2 + \frac{\angle B}{360^o} \times \pi r^2 + \frac{\angle C}{360^O} \times \pi r^2 + \frac{\angle D}{360^o} \times \pi r^2 \]
\[ = \left( \frac{\angle A + \angle B + \angle C + \angle D}{360^o} \right) \times \frac{22}{7} \times \left( 7 \right)^2 \]
\[ = \frac{360^o}{360^o} \times \frac{22}{7} \times 49 \left( \angle A + \angle B + \angle C + \angle D = 360^o\right)\]

∴ Area of the shaded region = 350 cm2 − 154 cm2 = 196 cm2
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2016-2017 (March) Foreign Set 3

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