Question

Plz solve it................​

Answer 1

Let the radii of the two circles be $\sf{r_1}$ and $\sf{r_2}$ where $\sf{r_1\ \textgreater\ r_2.}$

Since the two circles touch internally, the distance between their centers is the difference of their radii, i.e.,

$\sf{r_1-r_2=8\quad\longrightarrow\quad(1)}$

The sum of their areas is given as 130π cm², i.e.,

$\sf{\pi(r_1)^2+\pi(r_2)^2}=130\pi\\\\\\(r_1)^2+(r_2)^2=130\quad\longrightarrow\quad(2)$

Now, consider (1).

$\sf{r_1-r_2=8}$

Square both sides.

$\sf{(r_1-r_2)=8^2}\\\\\\\sf{(r_1)^2+(r_2)^2-2r_1r_2=64\quad\longrightarrow\quad(3)}$

But from (2),

$\sf{130-2r_1r_2=64}\\\\\\\sf{r_1r_2=33}$

Then, in (3),

$\sf{(r_1)^2+(r_2)^2-2r_1r_2+4r_1r_2=64+4\times33}\\\\\\\sf{(r_1)^2+(r_2)^2+2r_1r_2=196}\\\\\\\sf{(r_1+r_2)^2=14^2}\\\\\\\sf{r_1+r_2=14\quad\longrightarrow\quad(4)}$

On adding (1) and (4), we get,

$\large\text{ \underline{\underline{\mathsf{r_1=11\ cm}}} }$

And on taking difference between (1) and (4), we get,

$\large\text{ \underline{\underline{\mathsf{r_2=3\ cm}}} }$