Answer is 3:pi
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Explanation:
Check out the diagram below.
We have the following points
- A = center of both the circle and the square
- B = point on the circle directly to the left of point A
- P,Q,R,S = corner points of the square
- E = intersection of segments AB and PS
- F = point on the square directly to the right of point A
- C and D = intersection of segment PS and the circle
Let x be the length of segments DE and EC. Note how segment AB bisects segment DC. Whenever the radius is perpendicular to the chord like this, the radius will bisect the chord.
We're told that the chord length CD is the same as the radius AD, since CD = CE+ED = x+x = 2x, this means AD = 2x as well.
At this point, you probably can recognize we have a 30-60-90 triangle. One useful property of such triangles is that the hypotenuse is double the length of the short leg. The long leg is [tex]x\sqrt{3}[/tex] which is the length of segment EA. You can use the pythagorean theorem to confirm this.
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To summarize that last section, we formed triangle DAE and this triangle is a 30-60-90 triangle as shown in the diagram below.
Notice how the segments EA and FA span the entire width of the square. That width being [tex]EA+FA = x\sqrt{3}+x\sqrt{3} = 2x\sqrt{3} = EF[/tex]
The square has side lengths of [tex]2x\sqrt{3}[/tex] while the circle has radius 2x.
Use these two facts to find the area of each in terms of x.
[tex]A_1 = \text{area of square} = (\text{side})^2 = (2x\sqrt{3})^2 = 12x^2[/tex]
[tex]A_2 = \text{area of circle} = \pi*r^2 = \pi*(2x)^2 = 4\pi x^2[/tex]
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The last step is to divide the two results. The x^2 terms cancel out.
We'll be left with...
[tex]\frac{A_1}{A_2} = \frac{12x^2}{4\pi x^2} = \frac{12}{4\pi} = \frac{3}{\pi} \approx 0.9549[/tex]
There are many ways to express this ratio. One way we could write it is to say 3:pi to indicate that if the area of the square is 3, then the circle has area of pi (approximately 3.14). No matter what the square's area is, the circle is always slightly larger in area.
