Question

1. The radii of two circles are 14 cm and 7 cm respectively. Find the radius of the circle which
has circumference equal to the sum of the circumferences of the two circles. ​

Answer 1

Answer :

The required radius is 21 cm

Step-by-step explanation :

Given :

The radii of two circles are 14 cm and 7 cm respectively.

To find :

the radius of the circle which  has circumference equal to the sum of the circumferences of the two circles. ​

Solution :

First, let's find the circumference of each circle.

Let r₁ = 14 cm (radius of the first circle) and r₂ = 7 cm (radius of the second circle)

Circumference of the circle of radius 14 cm :

C₁ = 2πr₁

C₁ = 2π × 14

C₁ = 28π cm

Circumference of the circle of radius 14 cm :

C₂ = 2πr₂

C₂ = 2π × 7

C₂ = 14π cm

Let 'r cm' be the radius of the circle whose circumference is equal to the sum of the circumferences of the two circles.

Circumference of the circle of radius 'r' cm :

C = 2πr

Now,

The circumference of the required radius of the circle = Sum of the circumferences of the two circles

C = C₁ + C₂

2πr = 28π + 14π

 2r = 28 + 14

 2r = 42

 r = 42/2

 r = 21 cm

Therefore, the radius of the circle whose circumference equal to the sum of the circumferences of the two circles is 21 cm

Answer 2

☯︎ Given that,

• The radii of two circles are 14cm and 7cm respectively.

______________________

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

Let

• r1 = 14cm
• r2 = 7cm

Their circumferences as C1 and C2 respectively.

$\sf\underline{Firstly, \: find \: the \: perimeter \: of \: the \: first \: circle}$

$\boxed{\boxed{\sf\purple{circumference \: of \: the \: first \: circle \: = 2\pi \: r1}}}$

$\tt\implies \: c1 = 2 \times \pi \times 14$

$\tt\implies \: c1 = \boxed{\boxed{\sf\purple{28\pi \: cm}}}\bigstar$

$\sf\underline{Now, \: find \: the \: perimeter \: of \: the \: second \: circle}$

$\boxed{\boxed{\sf\purple{circumference \: of \: the \: second \: circle \: = 2\pi \: r2}}}$

$\tt\implies \: c2 = 2 \times \pi \times 7$

$\tt\implies \: c2 = \boxed{\boxed{\sf\purple{14\pi \: cm}}}\bigstar$

Let,

• Circumference of larger circle be C

$\sf\implies \: C = c1 + c2$

Now,

$\tt\implies \: C = 28π + 14π$

$\tt\implies \: 2\pi\: r = 28π + 14π$

$\tt\implies \: 2r = 28 + 14$

$\tt\implies \: 2r = 42$

$\tt\implies \: r = \cancel \dfrac {42}{2}$

$\tt\implies \: r = \boxed{\boxed{\sf\purple{21cm}}}\bigstar$

$\sf\therefore\underline{Hence, \: the \: radius \: of \: the larger \: circle \: is \: \bold{21cm}}$