# Ratio

• Dec 12th 2009, 05:02 AM
reiward
Ratio
We had this in a quiz, my answer was $\displaystyle \sqrt{2} = 1$

but my friends say the answer is 2 = 1 because they squared both sides. but I don't get why the need to square it. so here's the problem.

A square is inscribed in a circle, and another circle is inscribed in the square. What is the ratio of the area of the larger circle to the small circle?

What I did was

Let A1 = large circle
Let A2 = small circle

$\displaystyle R = \frac {\sqrt{2}}{2} e$

$\displaystyle r = \frac {1}{2} e$

$\displaystyle A1 = A2$

$\displaystyle \pi R^2 = \pi r^2$

I then cancel out pi and square leaving the radii.

$\displaystyle \frac {\sqrt{2}}{2} e= \frac {1}{2} e ] 2$

Thus becomes $\displaystyle \sqrt{2} = 1$
• Dec 12th 2009, 05:26 AM
Krahl
If you find a ratio you cant square it.
for example say im 3 times your height. so ratio is 3:1. square it 9:1 so now im 9 times your height.

Draw a circle, then draw a square within that circle then draw a circle within that square.

the radius R of the large circle is from the centre of the small circle to the corner of the square.

the radius r of the small circle is from the centre of the small circle to the side of the square. but then from that side of the square to the edge of the square should also be r.(since half the side of the square is r).

so we get a right angled triangle with hypoteneuse R and sides r.

so $\displaystyle R^2=r^2+r^2=2r^2$

$\displaystyle A1=(pi) R^2=2(pi)r^2$
$\displaystyle A2=(pi)r^2$

so ratio of 2:1
• Dec 12th 2009, 12:52 PM
Soroban
Hello, reiward!

Sorry, your friends are right . . .

Quote:

A square is inscribed in a circle, and another circle is inscribed in the square.
What is the ratio of the area of the larger circle to the small circle?

Consider one quadrant of the diagram.

Code:

      *       |    *       |        *  B     A * * - - - - *       |    *  * | *       |      *  |       |    *  * |  *       |  *      |    *       | *        *     O * - - - - - * - - *                   C

The radius of the large circle is: .$\displaystyle OB = R$
The area of the large circle is: .$\displaystyle A_1 \,=\,\pi R^2$

Since $\displaystyle \Delta OBC$ is a 45-45-90 right triangle: .$\displaystyle BC \,=\,\frac{R}{\sqrt{2}}$
The radius of the small circle is: .$\displaystyle r \,=\,\frac{R}{\sqrt{2}}$
The area of the small circle is: .$\displaystyle A_2 \:=\:\pi r^2 \:=\:\pi\left(\frac{R}{\sqrt{2}}\right)^2 \:=\:\frac{\pi R^2}{2}$

The ratio of areas is: .$\displaystyle \frac{A_1}{A_2} \;=\;\frac{\pi R^2}{\frac{\pi R^2}{2}} \;=\;2 \quad\Rightarrow\quad 2:1$
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