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A figure shows a trapezium $ABCD$ with two circles of equal diameter drawn in it. If $ON = 56 cm$ and $PM ⊥ BC, PM = 42 cm, AD = 30 cm$ and $BC = 64 cm$ . What is the area of the shaded region?

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722 c m^{2}

744 c m^{2}

742 c m^{2}

732 c m^{2}

answer is C.

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Detailed Solution

Given, a figure shows a trapezium

ABCD

with two circles of equal diameter drawn in it. Where,

ON = 56 cm

and

PM ⊥ BC, PM = 42 cm, AD = 30 cm

and

BC = 64 cm

Let the radius of each circle be

r .

From the figure, ON = 4r.
Then, 4r = 56
r = 14 cm
Area of a circle is

π r^{2}

.
Area of a trapezium =

\frac{1}{2}

(Sum of opposite sides)(Distance between them).
Area of shaded region

=

Area of trapezium

- 2

\times

Area of circle.

\Rightarrow

Area of shaded region

=

\frac{1}{2} \times (AD + BC) \times PM - 2 \times π r^{2}

\Rightarrow

Area of shaded region

=

\frac{1}{2} \times (30 + 64) \times 42 - 2 \times \frac{22}{7} \times 14 \times 14

\Rightarrow

Area of shaded region

= 1974 - 1232

\Rightarrow

Area of shaded region

= 742 {cm}^{2}

Hence, option 3 is correct.

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