Question

The area of the circle varies directly as the square of its radius. Find the radius of circle whose area is equal to the sum of the areas of two circles whose radii are 5 cm and 12 cm.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{A}\mathrm{denote}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{and}r\mathrm{denote}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}. \\ \therefore A\alpha r^2 \\ \Rightarrow A=k{r}^2(k\mathrm{is}\mathrm{a}\mathrm{constant}.)...\left(\mathrm{i}\right) \\ \mathrm{Let}R\mathrm{cm}\mathrm{be}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{whose}\mathrm{area}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{areas}\mathrm{of}\mathrm{two}\mathrm{circles}\mathrm{whose}\mathrm{radii}\mathrm{are}5\mathrm{cm}\mathrm{and}12\mathrm{cm}. \\ k{R}^2=k{\left(5\right)}^2+k{\left(12\right)}^2[\mathrm{From}(\mathrm{i}\left)\right] \\ \therefore R^2=25+144 \\ \Rightarrow R^2=169 \\ \Rightarrow R=13\mathrm{cm} \end{gathered}$$