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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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