The radii of two circles are 8 cm and 6 cm respectively.

Solution:

Let the radius of circles be $r \mathrm{~cm}, r_{1} \mathrm{~cm}$ and $r_{2} \mathrm{~cm}$ respectively. Then their areas are $A=\pi r^{2} \mathrm{~cm}^{2}, A_{1}=\pi r_{1}^{2} \mathrm{~cm}^{2}$ and $A_{2}=\pi r_{2}^{2} \mathrm{~cm}^{2}$ respectively.

It is given that,

Area $A$ of circle $=$ Area $A_{1}$ of circle $+$ Area $A_{2}$ of circle

$\pi r^{2}=\pi r_{1}^{2}+\pi r_{2}^{2}$

$\pi r^{2}=\pi\left(r_{1}^{2}+r_{2}^{2}\right)$

$r^{2}=r_{1}^{2}+r_{2}^{2}$

$r^{2}=r_{1}^{2}+r_{2}^{2}$

We have, $r_{1}=6 \mathrm{~cm}$ and $r_{2}=8 \mathrm{~cm}$

Substituting the values of $r_{1}, r_{2}$

$r^{2}=6 \times 6+8 \times 8$

$r^{2}=36+64$

$r^{2}=100$

$r=\sqrt{100}$

$r=10 \mathrm{~cm}$

Hence, the radius of circle is $10 \mathrm{~cm}$.