In the below figure, $ A B C D $ is a square of side $ 2 a $. Find the ratio between the circumferences.
"
Given:
\( A B C D \) is a square of side \( 2 a \).
To do:
We have to find the ratio between the circumferences.
Solution:
The square $ABCD$ is inscribed a circle.
Length of the side of the square $= 2a$
From the figure,
Diameter of the outer circle $AC =$ Diagonal of the square
$=\sqrt{2} \times 2 a$
$=2 \sqrt{2} a$
This implies,
Radius of the outer circle $R=\frac{\mathrm{AC}}{2}$
$=\frac{2 \sqrt{2} a}{2}$
$=\sqrt{2} a$
Diameter of the inner circle $=2a$
Radius of the inner circle $r=\frac{2a}{2}=a$
Therefore,
The ratio between the circumferences of the circles $=\frac{\text { circumference of outer circle }}{\text { circumference of inner circle }}$
$=\frac{2 \pi \mathrm{R}}{2 \pi r}$
$=\frac{\mathrm{R}}{r}$
$=\frac{\sqrt{2} a}{a}$
$=\frac{\sqrt{2}}{1}$
The ratio between the circumferences is $\sqrt{2}:1$.
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