Question 69

What is the radius of a circular field whose area is equal to the sum of the areas of three smaller circular fields of radii 8 m, 9 m and 12 m, respectively?

Solution

Let's assume the radius of a big circular field is 'R'.

Three smaller circular fields of radii 8 m, 9 m and 12 m, respectively.

$$r_1 = 8$$

$$r_2 = 9$$

$$r_3 = 12$$

area of big circular field = sum of the area of three smaller circular fields

$$\pi\ \times\ R^2 = \pi\ \times\ (r_1)^2+\pi\ \times\ (r_2)^2+\pi\ \times\ (r_3)^2$$

$$R^2=(r_1)^2+(r_2)^2+(r_3)^2$$

$$R^2=(8)^2+(9)^2+(12)^2$$

$$R^2=64+81+144$$
$$R^2=289$$
$$R^2=17^2$$
radius of a big circular field = R = 17 m

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