The radii of two circles are 8 cm and 6 cm. Find the radius of hte circle having area equal to the sum of the areas of the two circles.

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Solution

Let the radius of the first circle be $r_{1}$ , second circle be $r_{2}$ and third circle be R
Given,
$r_{1}$ =8cm; $r_{2}$ =6cm
According to the question;
$\Rightarrow$ $π R^{2} = π r_{1}^{2} + π r_{2}^{2}$
$\Rightarrow$ $π R^{2} = π (r_{1}^{2} + r_{2}^{2}$ )
$\Rightarrow$ $R^{2} = 64 + 36$
$\Rightarrow$ $R^{2} = 100$
$\Rightarrow$ R=10
Therefore, the radius of the third circle of 10cm.

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The radii of two circles are 8 cm and 6 cm. Find the radius of hte circle having area equal to the sum of the areas of the two circles.

Let the radius of the first circle be r1, second circle be r2 and third circle be R Given, r1 =8cm; r2 =6cm According to the question; ⇒ πR2=πr21+πr22 ⇒ πR2=π(r21+r22) ⇒ R2=64+36 ⇒ R2=100 ⇒ R=10 Therefore, the radius of the third circle of 10cm.