Question

The radii of two circles are 19 and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf {The\:radius\:of\:the\:1st\:circle = 19\: Cm\:(Given)}$

$\therefore\mathsf {Circumference\:of\:the\:1st\:circle = 2\pi \times 19 = 38\pi \:Cm}$

$\mathsf {The\:radius\:of\:the\:2nd\:circle = 9\: Cm\:(Given)}$

$\therefore\mathsf {Circumference\:of\:the\:2nd\:circle = 2\pi \times 19 = 18\pi \:Cm}$

SO,

$\mathsf {The\:sum\:of\:the\:circumference\:of\:two\:circles = 38\pi + 18\pi = 56\pi \: Cm}$

NOW,

$\mathsf {Let\:the\:radius\:of\:the\:3rd\:circle = R}$

$\therefore\mathsf {The\:circumference\:of\:the\:3rd\:circle = 2\pi\:R}$

It is given that sum of the circumference of two circles = Circumference of the 3rd circle.

$\implies \mathsf {Hence,56\pi = 2\pi\:R}$

$\implies \mathsf {Or,\:R = 28\:Cm}$

Answer 2

$\huge\bf{\underline{\red{Solution:-}}}$

$\bf{Given:-}\begin{cases}\sf{Radii\:of\:2\: circles=19cm\:and\:9cm}\end{cases}$

$\sf\:{\purple{We \:have\:To\:find\:the\:radius\:of\: circle}}\\ \sf whose\: circumference=sum\:of\:circumference \:of\:these\:2\: circle$

$\sf\:\:\:\:\:\:\:Now\:find\:the\: circumference\:of\: circles$

$\boxed{\bigstar{\sf{Circumference\:of\: circle=2\pi r}}}$

$\sf\implies Circumference\:if\: circle_1=2\times \pi \times 19\\ \\ \sf\implies Circumference\:of\: circle_1= 38 \pi$

$\sf\:\:\:\:\:\:\:\:\:Now\:the\: circumference\:of\: circle_2\\ \\ \sf\implies Circumference=2\times pi \times 9\\ \\ \sf\implies Circumference=18 \pi$

$\boxed{\sf{Circumference \:Of\: circle_1= 38 \pi }}$

$\boxed{\sf{Circumference \:Of\: circle_2= 18 \pi }}$

• So finally find the circumference of that circle whose radius we have to find

$\sf\bullet {\pink{Circumference=18\pi +38\pi }}\\ \\ \sf\bullet Circumference= 56\pi$

Now find the radius

$\sf\implies Circumference= 2\pi r\\ \\ \sf\implies 56 \cancel{\pi}=2\times \cancel{\pi}\times r\\ \\ \sf\implies 56=2r\\ \\ \sf\implies r=\cancel{\dfrac{56}{2}}\\ \\ \sf\implies r= 28$

$\boxed{\sf{\blue{Radius\:of\:new\: circle= 28cm}}}$