# Ex.12.1 Q1 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

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

## Text Solution

**What is known?**

Radii of two circles.

**What is unknown?**

Radius of \(3^\rm{rd}\) circle.

**Reasoning:**

Using the formula of circumference of circle \({C = 2}\pi r\) we find the radius of the circle.

**Steps:**

Radius \(({r_1})\)of \(1^ \rm{st}\) circle \(= 19 \,\rm{cm}\)

Radius \(({r_2})\) or \(2^\rm{nd} \) circle \(=9\,\rm{cm}\)

Let the radius of \(3^ \rm{rd}\) circle be \(r.\)

Circumference of \(1^\rm{st}\)circle \(= 2 \pi{{\text{r}}_{\text{1}}}{\text{ = 2 }}\pi (19) = 38 \pi\)

Circumference of\(2^\rm{ nd }\) circle \(= 2 \pi{{\text{r}}_{\text{2}}}{\text{ = 2 }}\pi (9) = 18 \pi\)

Circumference of \(3^\rm{rd}\)circle \(=2\pi r\)

Given that,

Circumference of \(3^\rm{rd}\) \(\rm{}circle\) \(=\) Circumference of \(1^\rm{st }\) \(\rm{}circle\) \(+\)Circumference of \(2^\text{nd}\)\(\rm{}circle.\)

\[\begin{align} 2 \pi r &=38 \pi+18 \pi \\ &=56 \pi \\ r &=\frac{56 \pi}{2 \pi} \\ &=28 \end{align}\]

Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is \(28\, \rm{cm.}\)