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detailed solution

Correct option is A

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 1st circle =  πr2$\mathrm{\pi }$$×$$×$ 8 = 64$\mathrm{\pi }$

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

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

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

Now, it is given that

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

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

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

r2  = 100

r =  $±$10

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

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detailed solution

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 1st circle =  πr2$\mathrm{\pi }$$×$$×$ 8 = 64$\mathrm{\pi }$

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

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

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

Now, it is given that

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

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

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

r2  = 100

r =  $±$10

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

+91

Are you a Sri Chaitanya student?