The diameter of a circle whose area is equal | vedclass

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Question

(D) Let the radii of the two circles be $r_{1} = 24 \, cm$ and $r_{2} = 7 \, cm$.The area of the first circle is $A_{1} = \pi r_{1}^{2} = \pi(24)^{2}

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Let the radii of the two circles be $r_{1} = 24 \, cm$ and $r_{2} = 7 \, cm$.The area of the first circle is $A_{1} = \pi r_{1}^{2} = \pi(24)^{2} = 576\pi \, cm^{2}$.The area of the second circle is $A_{2} = \pi r_{2}^{2} = \pi(7)^{2} = 49\pi \, cm^{2}$.Let the radius of the new circle be $R$. According to the problem,the area of the new circle is equal to the sum of the areas of the two circles:$\pi R^{2} = A_{1} + A_{2}$$\pi R^{2} = 576\pi + 49\pi$$\pi R^{2} = 625\pi$$R^{2} = 625$$R = \sqrt{625} = 25 \, cm$.The diameter of the new circle is $D = 2R = 2 	imes 25 = 50 \, cm$.

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii $24 \, cm$ and $7 \, cm$ is (in $cm$):

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