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.

Let us given that, 

Radius of the first circle = 8 cm

Radius of the second circle = 6 cm

We know that, Area of a circle = $\mathrm{\pi }$r²

Consider,

The radius of the third circle is r.

Area of 1^st circle =  πr² = $\mathrm{\pi }$$\times$8 $\times$ 8 = 64$\mathrm{\pi }$

Area of 2^nd circle = πr²=  $\mathrm{\pi }$$\times$6 $\times$ 6 = 36$\mathrm{\pi }$

Area of 3rd circle = $\mathrm{\pi }$r²

Circumference of 3^rd circle = 2π$\mathrm{\pi }$r 

Now, it is given that

Area of 3^rd circle = Area of 1^st circle  $+$ Area of 2^nd circle

$\mathrm{\pi }$r² =  64$\mathrm{\pi }$ $+$ 36$\mathrm{\pi }$

$\mathrm{\pi }$r²  = 100$\mathrm{\pi }$

r²  = 100

r =  $\pm$10

Hence, The radius of the circle having an area equal to the sum of the areas of the two circles is 10 cm.