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(A) Given,diameter of the first circle $d_{1} = 20 \, cm$,so its radius $r_{1} = \frac{20}{2} = 10 \, cm$.Diameter of the second circle $d_{2} = 48 \,

Vedclass · Accepted Answer

$52$

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(A) Given,diameter of the first circle $d_{1} = 20 \, cm$,so its radius $r_{1} = \frac{20}{2} = 10 \, cm$.Diameter of the second circle $d_{2} = 48 \, cm$,so its radius $r_{2} = \frac{48}{2} = 24 \, cm$.Sum of the areas of the two circles $= \pi r_{1}^{2} + \pi r_{2}^{2} = \pi(10)^{2} + \pi(24)^{2} = \pi(100 + 576) = 676\pi \, cm^{2}$.Let the radius of the new circle be $r$. Its area is $\pi r^{2}$.According to the problem,$\pi r^{2} = 676\pi$.$r^{2} = 676$.$r = \sqrt{676} = 26 \, cm$.Therefore,the diameter of the new circle $= 2r = 2 	imes 26 = 52 \, cm$.

Find the diameter of the circle whose area is equal to the sum of the areas of the two circles of diameters $20 \, cm$ and $48 \, cm$. (in $cm$)

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