The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
We have,
Radius of circle- $\mathrm{I}, \mathrm{r}_{1}=8 \mathrm{~cm}$
Radius of circle-II, $r_{2}=6 \mathrm{~cm}$
$\therefore \quad$ Area of circle- $\mathrm{I}=\pi \mathrm{r}_{1}^{2}=\pi(8)^{2} \mathrm{~cm}^{2}$
Area of circle- $\mathrm{II}=\pi \mathrm{r}_{2}^{2}=\pi(6)^{2} \mathrm{~cm}^{2}$
Let the radius of the circle-III be R
$\therefore \quad$ Area of circle-III $=\pi R^{2}$
$\Rightarrow \pi(8)^{2}+\pi(6)^{2}=\pi R^{2}$
$\Rightarrow \pi\left(8^{2}+6^{2}\right)=\pi R^{2}$
$\Rightarrow 8^{2}+6^{2}=R^{2}$
$\Rightarrow 64+36=R^{2}$
$\Rightarrow 100=\mathrm{R}^{2}$
$\Rightarrow 10^{2}=\mathrm{R}^{2} \Rightarrow \mathrm{R}=10 \mathrm{~cm}$
Thus, the radius of the new circle $=10 \mathrm{~cm}$.
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