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The diameter of a circular field is $ 40 \mathrm{~m} $ and that of another is $ 96 \mathrm{~m} $. Find the diameter of the circular field whose area is equal to the areas of two fields.
Given:
The diameter of a circular field is \( 40 \mathrm{~m} \) and that of another is \( 96 \mathrm{~m} \).
To do:
We have to find the diameter of the circular field whose area is equal to the areas of two fields.
Solution:
Diameter of circle 1 $=40\ m$
Radius of circle 1 $=\frac{40}{2}=20\ m$
Diameter of circle 2 $=96\ m$
Radius of circle 2 $=\frac{96}{2}=48\ m$
Area of circle 1 $=\pi (20)^2=400\pi$
Area of circle 2 $=\pi (48)^2=2304\pi$
Area of the circular field whose area is equal to the areas of two fields $=400\pi + 2304\pi$
$=2704\pi$
Let the radius of the circular field so formed be $r$.
This implies,
$\pi r^2=2704\pi$
$r^2=2704$
$r=\sqrt{2704}$
$r=52\ m$
Diameter $=2r=2(52)\ m=104\ m$
The diameter of the circular field whose area is equal to the areas of two fields is 104 m.
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