Question

The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm. Find the circumference of the circles.

Answer 1

Let the radii of the two circles be r₁ cm and r₂ cm.
Now,
Sum of the radii of the two circles = 7 cm
$r_{1+}{r}_2=7...\left(i\right)$

Difference of the circumferences of the two circles = 88 cm
$$\begin{gathered} \Rightarrow 2{\mathrm{\pi r}}_1-2{\mathrm{\pi r}}_2=8 \\ \Rightarrow 2\mathrm{\pi }({\mathrm{r}}_1-{\mathrm{r}}_2)=8 \\ \Rightarrow ({\mathrm{r}}_1-{\mathrm{r}}_2)=\frac{8}{2\mathrm{\pi }} \\ \Rightarrow {\mathrm{r}}_1-{\mathrm{r}}_2=\frac{8}{2\times {\displaystyle \frac{22}{7}}} \\ \Rightarrow {\mathrm{r}}_{1-}{\mathrm{r}}_2=\frac{8\times 7}{44} \end{gathered}$$

$$\begin{gathered} r_1-r_2=\frac{56}{44} \\ {r}_1-r_2=\frac{14}{11}...\left(ii\right) \end{gathered}$$

Adding (i) and (ii), we get:
$$\begin{gathered} 2r_1=\frac{91}{11} \\ {r}_1=\frac{91}{22} \end{gathered}$$

∴ Circumference of the first circle = $2{\mathrm{\pi r}}_1$
$$\begin{gathered} =2\times \frac{22}{7}\times \frac{91}{22} \\ =26\mathrm{cm} \end{gathered}$$
Also,
$$\begin{gathered} r_1-r_2=\frac{14}{11} \\ \frac{91}{22}-r_2=\frac{14}{11} \\ \frac{91}{22}-\frac{14}{11}=r_2 \\ {r}_2=\frac{63}{22} \end{gathered}$$

∴ Circumference of the second circle = $2{\mathrm{\pi r}}_2$

$$\begin{gathered} =2\times \frac{22}{7}\times \frac{63}{22} \\ =18\mathrm{cm} \end{gathered}$$

Therefore, circumferences of the first and second circles are 18 cm and 26 cm, respectively.