Question

What is the diameter of a circle whose area is equal to the sum of the areas of two circles of radii 40cm and 9cm ?

Answer 1

Hint: Proceed by finding the area of both the circles and adding their areas to find the area of the big circle. Assume the radius of the big circle and substitute the area in the equation of the formula to get to the desired answer.

Complete step-by-step solution -

The area of circle with radius 40cm = $\pi \times {40^2}c{m^2}$ ( since area of the circle = $\pi {r^2}$ )
The area of circle with radius 9cm = $\pi \times {9^2}c{m^2}$
Now we know that the area of the big circle is equal to the sum of the areas of two circles of radii 40 cm and 9cm.
Therefore the sum of areas of two small circles is $\pi \times {40^2}c{m^2}$+ $\pi \times {9^2}c{m^2}$= $\pi \left( {{{40}^2} + {8^2}} \right)c{m^2}$
$ = \pi \times 1681c{m^2}$
$ = \pi \times {41^2}c{m^2}$
But area is equal to the area of the big circle ,
Hence area of big circle = $ = \pi \times {41^2}c{m^2}$
$ \Rightarrow \pi \times Radiu{s_{big}}^2 = \pi \times {41^2}$
On comparing ,
Radius of big circle = 41 cm
Therefore diameter = 2R = $2 \times 41 = 81cm$

Note: Remember that in such types of questions we need to recall the formula of area of a circle to solve the question. Note that while calculating areas of small circles we need not to solve them completely as some part of it would get cancelled out when we equate it with the area of the big circle.