Question

The radii of two circles are 19 cm and 9cm respectively find the radius of the circle which has a circumference equal to the sum of the circumferences of the two given circles

Answer 1

$\huge\star{\red{\underline{\mathfrak{Answer :-}}}}$

28 cm

$\huge\star{\red{\underline{\mathfrak{Explanation :-}}}}$

Given :-

The radii of 1st circle = 19 cm

The radii of 2nd circle = 9 cm

The sum of circumference of both the circles is equal to the circumference of 3rd circle.

To find :-

The radius of the third circle.

Solution :-

The circumference of a circle is =$2\pi \times r$

So,

Circumference of 1st circle =$2\pi \times 19$

Circumference of 2nd circle =$2\pi \times 9$

Circumference of 3rd circle =$2\pi \times r$

Therefore,

The sum of the circumference of both the circles equals the circumference of the third circle.

$\implies(2\pi \times 19) + (2\pi \times 9) = (2\pi \times r)$

$\implies \cancel {2\pi} \times (19 + 9) = \cancel {2\pi} \times r$

$\implies r = 19 + 9$

$\implies r = 28 \: cm$

___________________________________

Answer 2

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Circle 1

• Radius of the circle is 9 cm

Circumference

$\sf{c = 2\pi \: r} \\ \\ \rightarrow \: \sf{c {}_{1} = 2\pi \: (9)} \\ \\ \rightarrow \: \boxed{\sf{ {c}_{1} = 18 \: \pi \: {cm}^{2} }}$

Circle 2

• Radius is 19 cm

Circumference

$\sf{ {c}_{2} = 2 \pi \: (18)} \\ \\ \rightarrow \: \boxed{\sf{ {c}_{2} = 38 \pi \: {cm}^{2} }}$

Given

Sum of the circumferences of these two circles is equal to a bigger circle.

Let R be the radius of bigger circle

Implies,

$\sf{ {c}_{1} + {c}_{2} = C} \\ \\ \leadsto \: \sf{18\pi + 38\pi = 2\pi \: R} \\ \\ \leadsto \: \sf{2\pi \: R = 56\pi} \\ \\ \huge{ \leadsto \: \tt{R = 28 \: cm}}$