What is the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24cm and 7cm?

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Solution

We have the formula for area of circle,

$A r e a = π r^{2}$ , where $r$ is the radius of the circle

Let, the circle whose area is equal to the sum of the areas of the two circles of radii $24 c m$ and $7 c m$ be $A_{3}$ and its radius be $r_{3}$

Area of circle with radius, $r_{1} = 24 c m$ be $A_{1}$

Area of circle with radius, $r_{2} = 7 c m$ be $A_{2}$

Given,

$A_{3} = A_{1} + A_{2}$

$π (r_{3})^{2} = π (r_{1})^{2} + π (r_{2})^{2}$

$π (r_{3})^{2} = π [(r_{1})^{2} + (r_{2})^{2}]$

$(r_{3})^{2} = (r_{1})^{2} + (r_{2})^{2}$

$(r_{3})^{2} = (24)^{2} + (7)^{2} = 625 = 25^{2}$

$r_{3} = 25$

$D i a m e t e r = 2 \times r a d i u s = 2 \times 25 = 50 c m$

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