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July 1st, 2016, 03:03 PM   #1
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Finding the area of a circle inside a square only given the area of the square

How to determine the area of a circle/eclipse inscribed inside a square/rectangle (that perfectly fits in the square) if the only information provided is the area of the square?

Example:

The area of a rectangle is 45, the length and height is impossible to determine, what is the area of the circle that is inside the square.
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July 1st, 2016, 03:59 PM   #2
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You need the dimensions to determine the radius of the circle, so the problem is solvable only for a square.
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July 1st, 2016, 04:52 PM   #3
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Hi v8archie,

The question I posed in the original post is solvable, very easily actually but I'm having trouble finding anything on the internet about it. Which means either I created a new formula or it's hard to search for or just no one cares. I'm assuming that it is just that it's hard to search which is why I posted the question here hoping some mathmatician will see it and clear it up for me
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July 1st, 2016, 04:58 PM   #4
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I could also present the question like this:

Given a unit square with a area of 36 (which therefore must have a length and height of 6) solve for the area of the unit circle inscribed in the circle without using the radius, diameter, length or height in your calculation. I presented the question more complicated to begin with just to make sure it was understood that the radius/diameter should not be used since it's not given and can't be determined.
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July 2nd, 2016, 05:29 PM   #5
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Quote:
Originally Posted by skimanner View Post
I could also present the question like this:

Given a unit square with a area of 36 (which therefore must have a length and height of 6) solve for the area of the unit circle inscribed in the circle without using the radius, diameter, length or height in your calculation. I presented the question more complicated to begin with just to make sure it was understood that the radius/diameter should not be used since it's not given and can't be determined.
A "unit square" is a 1 by 1 square, thus area 1.
Why are you bringing in the term "unit"?

If square is 6 by 6 (area 36), then inscribed circle
has area pi * (6/2)^2 = pi * 9

However, that's as basic as it gets; so you must be
asking something else: WHAT is it?
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July 4th, 2016, 01:25 AM   #6
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Let $\displaystyle r$ be the radius of a circle and $\displaystyle a$ be the side length of the square.

If the circle is contained inside the square, then the centres of both shapes are at the same location and

$\displaystyle r = \frac{a}{2}$

Therefore, the area of the circle, $\displaystyle A_c$, is

$\displaystyle A_c = \frac{\pi a^2}{4}$

Since the area of the square, $\displaystyle A_s$, is $\displaystyle a^2$,

$\displaystyle A_c = \frac{\pi A_s}{4}$


Since $\displaystyle \frac{\pi}{4} = 0.7854$ to 4 d.p., there is always a reduction in area of about 78.54%.

Last edited by Benit13; July 4th, 2016 at 01:27 AM.
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July 4th, 2016, 05:49 PM   #7
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The original question posed is not perfectly posed.

$A = area\ of\ square \implies \sqrt{A} = length\ of\ any\ side.$

So the radius of an inscribed circle is $\dfrac{\sqrt{A}}{2}$

And the area of that circle is $\dfrac{\pi A}{4}$ as benit showed.

I think that the original question correctly asserts that the area of an ellipse inscribed in a rectangle of given area is similarly determinable without any knowledge of the rectangle's linear dimensions.

Area of rectangle $= B.$

Unknown length of rectangle $= w \implies$

Unknown height of rectangle $ = h = \dfrac{B}{w}.$

Area of inscribed ellipse $= \pi * \dfrac{h}{2} * \dfrac{w}{2} = \dfrac{\pi * B * \cancel{w}}{2 * 2 * \cancel{w}} = \dfrac{\pi B}{4}.$

In other words it was an assertion given without demonstration, not a question at all.

Last edited by JeffM1; July 4th, 2016 at 05:52 PM.
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July 5th, 2016, 05:22 PM   #8
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Thanks benit and Jeff. Area of square/rectangle x pi/4 = area of circle/ellipse was what I had figured out, thanks for help
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July 5th, 2016, 05:34 PM   #9
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By the way do either of you know why all of the regular pentagon area formulas seem overly complicated? From what I have found the area is side squared x ~1.72048 which I'm sure could be contained in a better way such as 1 + sqrt(2) that is used for a regular octagon.
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July 5th, 2016, 06:22 PM   #10
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https://www.algebra.com/algebra/home...on.382616.html
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