Discussion
The perimeters of two circular fields C1 and C2 are 40 π m and 96 π 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 ?
Answer: A
The radius of C3 is 52 m.
A perimeter of \(2\pi r\) gives the radii: \(2\pi r_1 = 40\pi\), so \(r_1 = 20\) m; \(2\pi r_2 = 96\pi\), so \(r_2 = 48\) m.
\[ \begin{aligned} \pi r_3^2 &= \pi(20^2 + 48^2) \\ r_3^2 &= 400 + 2304 = 2704 \\ r_3 &= 52\ \text{m} \end{aligned} \]
Adding the radii (68 m) instead of the areas is the trap. 20, 48, 52 is 4 times the triple 5, 12, 13, which is a quick check.
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