Question

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 $12m,9m$ and $8m$ respectively?

• A. $29m$
• B. $17m$
• C. $13m$
• D. $12m$

Answer 1

Answer: (B)

$17m$

We know that,

Area of circle $=\pi {\left(radius\right)}^2$.

Let $r$ be the radius of the circle.

$\therefore$ Area of circle $=\pi r^2$.

Area of circle of radius $12m=\pi {\left(12\right)}^2=144\pi$.

Area of circle of radius $9m=\pi {\left(9\right)}^2=81\pi$.

Area of circle of radius $8m=\pi {\left(8\right)}^2=64\pi$.

Given that the area of circle is equal to the sum of the areas of the three smaller circular fields.

$\therefore \pi r^2=144\pi +81\pi +72\pi$

$\Rightarrow \pi r^2=289\pi$

$\Rightarrow r=\sqrt{289}=17m$.

Hence the radius of the required circular field is $17m$.

Therefore, option $B$ is correct.