Question

Two circles touch each other externally. The sum of their areas is 490 π cm². Their centres are separated by 28 cm. The difference of their perimeters is ___________.

Answer 1

Let R and r be the radii of the two circles. Suppose R > r.

We know that if two circles touch each other externally, then the distance between their centres is equal to the sum of the radii of the two circles.

∴ R + r = 28 cm .....(1)

Also,

Sum of the areas of the two circles = $490\mathrm{\pi }$ cm²

$$\begin{gathered} \Rightarrow \mathrm{\pi }{R}^2+\mathrm{\pi }{r}^2=490\mathrm{\pi } \\ \Rightarrow R^2+r^2=490.....\left(2\right) \end{gathered}$$

Now,

$$\begin{gathered} {\left(R+r\right)}^2+{\left(R-r\right)}^2=2\left(R^2+r^2\right) \\ \Rightarrow {\left(28\right)}^2+{\left(R-r\right)}^2=2\times 490\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right] \\ \Rightarrow {\left(R-r\right)}^2=980-784=196 \\ \Rightarrow R-r=\sqrt{196}=14\mathrm{cm}.....\left(3\right) \end{gathered}$$

∴ Difference between their perimeters

$$\begin{gathered} =2\mathrm{\pi }R-2\mathrm{\pi }r \\ =2\mathrm{\pi }\left(R-r\right) \\ =2\times \frac{22}{7}\times 14\left[\mathrm{Using}\left(3\right)\right] \\ =88\mathrm{cm} \end{gathered}$$

Two circles touch each other externally. The sum of their areas is 490 π cm². Their centres are separated by 28 cm. The difference of their perimeters is _____88 cm______.