The perimeters of two circular fields C₁ and C₂ are 40π m and 96π m, respectively. What is the radius of the third circular field C₃ whose area is equal to the sum of the areas of C₁ and C₂ ?

The correct answer is 52 m.

Let the radii of the two circular fields C₁ and C₂ be r₁ and r₂, respectively.

Step 1: Find the radii r₁ and r₂ using their perimeters.

The perimeter (circumference) of a circle is given by the formula:

$\text{Perimeter}=2\pi r$

For field C₁, the perimeter is 40π m:

$2\pi r_1=40\pi$

$r_1=20\text{ m}$

For field C₂, the perimeter is 96π m:

$2\pi r_2=96\pi$

$r_2=48\text{ m}$

Step 2: Calculate the areas of fields C₁ and C₂.

The area of a circle is given by the formula:

$\text{Area}=\pi r^2$

Area of C₁ (A₁):

$A_1=\pi {\left(20\right)}^2=400\pi \text{ m}{}^2$

Area of C₂ (A₂):

$A_2=\pi {\left(48\right)}^2=2304\pi \text{ m}{}^2$

Step 3: Find the radius of field C₃.

Let R be the radius of field C₃. The area of C₃ (A₃) is equal to the sum of the areas of C₁ and C₂:

$A_3=A_1+A_2$

$\pi R^2=400\pi +2304\pi$

$\pi R^2=2704\pi$

Dividing both sides by π:

$R^2=2704$

$R=\sqrt{2704}=52\text{ m}$

Thus, the radius of the third circular field C₃ is 52 m.