Questions

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.

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 = $\mathrm{\pi}$r^{2}

Consider,

The radius of the third circle is r.

Area of 1^{st} circle = πr^{2} = $\mathrm{\pi}$$\times $8 $\times $ 8 = 64$\mathrm{\pi}$

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

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

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

$\mathrm{\pi}$r^{2} = 64$\mathrm{\pi}$ $+$ 36$\mathrm{\pi}$

$\mathrm{\pi}$r^{2} = 100$\mathrm{\pi}$

r^{2} = 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.

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detailed solution

Correct answer is 1

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 = $\mathrm{\pi}$r^{2}

Consider,

The radius of the third circle is r.

Area of 1^{st} circle = πr^{2} = $\mathrm{\pi}$$\times $8 $\times $ 8 = 64$\mathrm{\pi}$

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

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

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

$\mathrm{\pi}$r^{2} = 64$\mathrm{\pi}$ $+$ 36$\mathrm{\pi}$

$\mathrm{\pi}$r^{2} = 100$\mathrm{\pi}$

r^{2} = 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.

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