Question

the radii of two circles are 19and 9respectively find the radius of the circle which has circumfrence equal to the sum of the circumfrence of the two circles​

Answer 1

Given

Radii of two circles are 9 and 19 units

Let the circle with radius 9 units be circle 1 and the one with 19 units radius be 2nd circle

Circle 1:

Radius = 9 units

$\boxed{ \sf \: circumference \: of \: a \: circle \: = 2\pi \: r \: units}$

$P_1=2\pi (9)$

$P_1=18\pi \: units$

Circle 2:

radius =19 units

$P_2=2\pi(19)$

$P_2=38\pi \: units$

Circle 3:

let radius be r

circumfrence equal to the sum of the circumfrence of the two circles

$P=P_1+P_2$

$2\pi r=18\pi+38\pi$

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

cancelling π on both sides

$2r=56$

$r=\dfrac{56}{2}$

$\boxed{\sf r=28 \: units }$

$\sf required \:radius \: is \:28 \: units$

Answer 2

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According to the given question:-

• The radii of two circles are 19 and 9 respectively.

• We have to find the radius of the circle which has circumfrence equal to the sum of the circumfrence of the two circles.

• Let one radii (9) be - (1)
• Let other radii (19) be - (2)

So,

$\boxed\bf{[2 \:π r ⟶ circumfrence \:of \:circle]}$

Circle - (1)

$\sf⟶9 \: is \: the \: radii$

$\sf⟶p _{1} = 2\pi(9)$

$\sf⟶p _{1} = 2\pi \: = \: units$

Circle - (2)

$\sf⟶19 \: units \: is \: the \: radii$

$\sf⟶p _{2} = 2\pi(19)$

$\sf⟶p _{2} = 38\pi \: = units$

Circle - (3)

• Let (r) be the radii => Radius.

So,

$\sf⟶p = p _{1} + p _{2}$

$\sf⟶p\pi \: r \: = > 18\pi + 38\pi$

$\sf⟶2\pi \: r \: = > \pi(18 + 38)$

Cancelling pi (π) for sides

$\sf⟶2r = 56$

$\sf⟶r = \frac{56}{2} = 28$

$Therefore, \: r \: = 28 \: units.$