Question

The diameter of the two circles is 42 cm, respectively. And 56 cm. . Determine the radius of the third circle if the area of ​​another third circle is equal to the sum of the areas of the first two circles.​

Answer 1

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{The \ radius \ of \ third \ circle \ is \ 35 \ cm}$

$\sf\orange{Given:}$

$\sf{\implies{Diameter \ of \ two \ circles \ are \ 42 \ cm}}$

$\sf{and \ 56 \ cm.}$

$\sf{\implies{Area \ of \ third \ circle \ is \ equal}}$

$\sf{to \ sum \ of \ area \ of \ first \ two \ circles.}$

$\sf\pink{To \ find:}$

$\sf{Radius \ of \ the \ third \ circle.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Radius \ of \ first \ two \ circles \ are}$

$\sf{\frac{42}{2} \ cm \ and \ \frac{56}{2} \ cm \ respectively.}$

$\sf{i.e. \ 21 \ cm \ and \ 28 \ cm \ respectively.}$

____________________________________

$\sf{Area \ of \ circle=\pi\times \ r^{2}}$

$\sf{...formula}$

$\sf{Let \ the \ radius \ of \ third \ circle \ be \ R}$

$\sf{Area \ of \ third \ circle=}$

$\sf{Area \ of \ first \ circle + Area \ of \ second \ circle}$

$\sf{\pi\times \ R^{2}=\pi\times21^{2}+\pi\times28^{2}}$

$\sf{\pi\times \ R^{2}=\pi\times(21^{2}+28^{2})}$

$\sf{\therefore{R^{2}=21^{2}+28^{2}}}$

$\sf{R^{2}=441+784}$

$\sf{R^{2}=1225}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{R=35 \ cm}$

$\sf\purple{\tt{\therefore{The \ radius \ of \ third \ circle \ is \ 35 \ cm}}}$