Home/Class 10 Math Chapter List/14. Areas Related to Circles/# The radii of two circles are(19;cm) and (9;cm)respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles(.)

The radii of two circles are\(19\;cm\) and \(9\;cm\)respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles\(.\)

28 cm

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Let \(r_{1}=19\;cm\) and \(r_{2}=9\;cm\)and r be the radius of new circle\(.\)

\(\therefore \)Circumference of \(1^{st}\) circle\(,\)

\(C_{1}=2\pi r_{1}=2\pi \times 19=38\pi\; cm\)

and circumference of \(2^{nd}\) circle\(,\)

\(C_{2}=2\pi r_{2}=2\times \pi \times 9=18\pi\; cm\)

According to the given condition\(,\)

Circumference of new circle \(=\) Circumference of \(1^{st}\) circle\(+\) Circumference of \(2^{nd}\) circle

\(\therefore \) \(2\pi r=38\pi +18\pi \)

\(\Rightarrow 2\pi r=56\pi \)

\(\Rightarrow\)\(r=\frac {56\pi }{2\pi }=28\;cm\)

Hence\(,\) radius of new circle is\(28cm.\)

\(\therefore \)Circumference of \(1^{st}\) circle\(,\)

\(C_{1}=2\pi r_{1}=2\pi \times 19=38\pi\; cm\)

and circumference of \(2^{nd}\) circle\(,\)

\(C_{2}=2\pi r_{2}=2\times \pi \times 9=18\pi\; cm\)

According to the given condition\(,\)

Circumference of new circle \(=\) Circumference of \(1^{st}\) circle\(+\) Circumference of \(2^{nd}\) circle

\(\therefore \) \(2\pi r=38\pi +18\pi \)

\(\Rightarrow 2\pi r=56\pi \)

\(\Rightarrow\)\(r=\frac {56\pi }{2\pi }=28\;cm\)

Hence\(,\) radius of new circle is\(28cm.\)

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