Question

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.
​

Answer 1

let the radius be r1 and r2 of two circles and the radius of 3rd circle be r

A. T. Q,

circumference = circumference of 1st circle + circumference of 2nd circle

2πr = 2π r1 + 2πr2

2πr = 2π(r1+r2)

r= r1 + r2

r = 19 + 9

r= 28

Answer 2

Step-by-step explanation:

$\frak{ Given }\begin{cases}\sf{ Radius \ of \ 1st \ circle = 19 \ cm \:}\\\sf{ Radius \ of \ 2nd \ circle = 9 \ cm \:}\end{cases}$

We've to find out the radius of the circle which has circumference equal to the sum of the circumference of the two circle.

$\underline{\:\large{\textit{1. \sf Circumference of 1st circle :}}}$

$\star \ \boxed{\sf{\purple{ Circumference \: = \: 2 \pi r}}}$

$:\implies\sf Circumference = 2 \pi \Big( 19 \Big) \\\\\\:\implies\boxed{\frak{\pink{ \: 38 \pi \: }}}$

$\underline{\:\large{\textit{1. \sf Circumference of 2nd circle :}}}$

$:\implies\sf Circumference = 2 \pi \Big( 9 \Big) \\\\\\:\implies\boxed{\frak{\pink{\: 18 \pi \; }}}$

Circumference of Both the circles is 38π & 18π.

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$\star$ Circumference of third Circle = Circumference of 1st circle + circumference of 2nd circle.

$:\implies\sf 2 \pi r = 38 \pi + 18 \pi \\\\\\:\implies\sf 2 \pi r = 56 \pi \\\\\\:\implies\sf r = \cancel\dfrac{56 \pi}{ 2 \pi}\\\\\\:\implies\underline{\boxed{\frak r = 28}}$

$\therefore\:\underline{\textsf{Hence, required radius is \textbf{28 cm}}}.$