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July 1st, 2016, 03:03 PM  #1 
Newbie Joined: Aug 2009 Posts: 12 Thanks: 0  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. 
July 1st, 2016, 03:59 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
You need the dimensions to determine the radius of the circle, so the problem is solvable only for a square.

July 1st, 2016, 04:52 PM  #3 
Newbie Joined: Aug 2009 Posts: 12 Thanks: 0 
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 
July 1st, 2016, 04:58 PM  #4 
Newbie Joined: Aug 2009 Posts: 12 Thanks: 0 
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. 
July 2nd, 2016, 05:29 PM  #5  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Quote:
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?  
July 4th, 2016, 01:25 AM  #6 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
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. 
July 4th, 2016, 05:49 PM  #7 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
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. 
July 5th, 2016, 05:22 PM  #8 
Newbie Joined: Aug 2009 Posts: 12 Thanks: 0 
Thanks benit and Jeff. Area of square/rectangle x pi/4 = area of circle/ellipse was what I had figured out, thanks for help

July 5th, 2016, 05:34 PM  #9 
Newbie Joined: Aug 2009 Posts: 12 Thanks: 0 
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.

July 5th, 2016, 06:22 PM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  

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area, circle, finding, inside, square 
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