Find the radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters $36\ cm$ and $20\ cm$.


Given: A circle whose circumference is equal to the sum of the circumferences of the two circles of diameters $36\ cm$ and $20\ cm$.

To do: To find the radius of the $3^{rd}$ circle.

Solution:

As given, diameters of the circle $d_1=36\ cm$ and $d_2=20\ cm$.

Let radius of one circle $r_1=\frac{d_1}{2}=\frac{36}{2}=18\ cm$

Radius of $2^{nd}$ circle $r_2=\frac{d_2}{2}=\frac{20}{2}=10\ cm$

Radius of $3^{rd}$ circle $r_3=?$

Acording to the question:

$C_1+C_2=C_3$

$2\pi r_1 +2\pi r_2 =2\pi r_3$

$2\pi  (r_1 + r_2) = 2\pi r_3$

$18 + 10 = \frac{2\pi r_3}{2\pi}$

$28 = r_3$

Thus, radius of $3^{rd}$ circle is $28\ cm$.

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Updated on: 10-Oct-2022

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