The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circles which has it circumference equal to the sum of the circumferences of the two circles.
Let the radius of circles be $r \mathrm{~cm}, r_{1} \mathrm{~cm}$ and $r_{2} \mathrm{~cm}$ respectively. Then their circumferences are $C=2 \pi r \mathrm{~cm}, C_{1}=2 \pi r_{1} \mathrm{~cm}$ and $C_{2}=2 \pi r_{2} \mathrm{~cm}$ respectively.
It is given that,
Circumference $C$ of circle = Circumference $C_{1}$ of circle+Circumference $C_{2}$ of circle
$2 \pi r=2 \pi r_{1}+2 \pi r_{2}$
$2 \pi r=2 \pi\left(r_{1}+r_{2}\right)$
$r=r_{1}+r_{2}$
We have, $r_{1}=19 \mathrm{~cm}$ and $r_{2}=9 \mathrm{~cm}$
Substituting the values of $r_{1}, r_{2}$
$r=19+9$
$r=28 \mathrm{~cm}$
Hence the radius of the circle is $28 \mathrm{~cm}$.
We know that the area A of circle is
$A=\pi r^{2}$
Substituting the value of r
$A=\frac{22}{7} \times 28 \times 28$
$=2464 \mathrm{~cm}^{2}$
Hence the area of the circle is $2464 \mathrm{~cm}^{2}$.