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The circumference of a circular field is 396 m and that of the other circular field is 132 m. Find the area (in m2^2) of the third circular field whose radius is the sum of the radii of the first two fields. Take π=227\pi = \frac{22}{7}

Solution

Correct Option: 3

From 2πr1=3962\pi r_1 = 396, r1=63r_1 = 63 m. From 2πr2=1322\pi r_2 = 132, r2=21r_2 = 21 m.

New radius =63+21=84= 63 + 21 = 84 m.

Area =πr2=227×84×84=22176= \pi r^2 = \frac{22}{7} \times 84 \times 84 = 22176 m2^2.

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