Question

The radius of one circluar field is 20 m and that of another is 48 m. Find the radius of the third circular field whose area is equal to the sum of the areas of two fields.

Answer 1

Let the area of the circle whose radius is 20 m be A₁ , and the area of the circle whose radius
is 48 m be A₂. Let A₃ be the area of a circle that is equal to the sum of the areas of the two fields, with the radius of its field being r cm.
∴ A₃ = A₁ + A₂

$$\begin{gathered} A_1=\mathrm{\pi }{\left(20\right)}^2=\left(\frac{22}{7}\times 20\times 20\right){\mathrm{m}}^2=\left(400\mathrm{\pi }\right){\mathrm{m}}^2 \\ {A}_2=\mathrm{\pi }(48{)}^2=\left(\frac{22}{7}\times 48\times 48\right){\mathrm{m}}^2=\left(2304\mathrm{\pi }\right){\mathrm{m}}^2 \\ {A}_3=A_1+A_2=\left(400\mathrm{\pi }\right)+\left(2304\mathrm{\pi }\right)=\mathrm{\pi }\left(400+2304\right){\mathrm{m}}^2 \\ \Rightarrow A_3=\mathrm{\pi }{\left(r\right)}^2=\mathrm{\pi }\left(400+2304\right){\mathrm{m}}^2 \\ \Rightarrow {\left(r\right)}^2=\left(400+2304\right){\mathrm{m}}^2 \\ \Rightarrow r=\sqrt{2704}\mathrm{m}=52\mathrm{m} \end{gathered}$$