Question

Let C and A be the circumference and area of a circle respectively. If xC is the circumference of the another circle whose area is \[2A,\] then x equals
A) $ 2\sqrt 2 $
B) $ 2 $
C) $ \sqrt 2 $
D) $ \dfrac{1}{2} $

Answer 1

Hint: To solve this question, i.e., to find the value of x. First, we will equate the areas of two circles with different radii, here we will get the relation between the two radii of the circles. After that we will equate the circumference of both the circles, and then using the relation of both the radii in the formula, we will get the value of x.

Complete step-by-step answer:
We have been given that C and A are the circumference and area of the circle respectively. And, xC is the circumference of the other circle whose area is \[2A,\] we need to find the value of x.
Let us draw a figure of two circles, to understand better.

We know that, the area of the circle $ = \pi {r^2}. $
So, there are two circles, the area of first circle of radius $ {r_1} = A = \pi {r_1}^2 $
And, the area of another circle of radius $ {r_2} = 2A = 2(\pi {r_2}^2) $
On equating both the areas of the circle, we get
 $ \pi {r_1}^2 = 2\pi {r_2}^2 $
 $ {r_1} = \sqrt 2 {r_2}.......eq.(1) $
Now, we know that, circumference of a circle $ = 2\pi r. $
So, the circumference of first circle of radius \[{r_1} = {\text{ }}C = 2\pi {r_1}\] $ $
And, the circumference of another circle of radius \[{r_2} = xC = x(2\pi {r_2})\]
On equating both the circumference of the circle, we get
\[2\pi {r_1} = x2\pi {r_2}\]
Now, on putting the value of $ {r_1} = \sqrt 2 {r_2} $ from \[eq.\left( 1 \right),\] we get
\[2\pi (\sqrt 2 {r_2}) = x2\pi {r_2}\]
\[ \Rightarrow x = \sqrt 2 \]
So, the value of x is $ \sqrt 2 $.
So, the correct answer is “Option C”.

Note: Students should note that these types of questions always take the different values of radius, because if you choose the radius of both the circles the same, you will get confused and end up getting your answer incorrect. So, always keep in mind to assume the value of both the radius differently.