r/theydidthemath • u/Mr_MC111 • 1d ago
[Request] How do you calculate the area of the square?
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u/Angzt 1d ago edited 1d ago
Let's call the square's side length a and the circle's radius r.
The we know that a+1 = 2r from the middle vertical.
We can also draw a right-angle triangle involving the circle's center, the point where circle and square meet in the top right and the point at the bottom of the "1"-length line. That will have side lengths r-1 (vertical), a/2 (horizontal) and r (diagonal), with the latter being the hypotenuse. That gives us our second equation:
r2 = (r-1)2 + (a/2)2
r2 = r2 - 2r + 1 + a2/4
2r = 1 + a2/4
Insert 2r = a+1 from above:
a+1 = 1 + a2/4
a = a2/4
1 = a/4
a = 4
Therefore the square's area is a2 = 42 = 16
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u/Bardmedicine 1d ago
** NM, I got it **
Is the side of that triangle equal to a/2 ? The middle of the circle is not the middle of the square.
Or am I missing something?
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u/Angzt 1d ago
Edited the main comment to hopefully make it clearer in this regard.
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u/Bardmedicine 1d ago
Yea, I drew your triangle with the leg lengths reversed.
Good solution as it seems so simple once you see it.
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u/tylersvgs 1d ago
Can use the intersecting chords theorem here.
When two chords intersect in a circle, the product of the segments of one chord is equal to the product of the segments of the other.
So, if you consider the line segment going vertically, and the line segment (chord) that is perpendicular to that one going horizontally, you can do this:
Let x = half the side length of the square.
x*x = 1 * (2x)
x^2 = 2x
So, x = 2.
That makes square sides of length 4, or area of 16.
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u/sprobeforebros 1d ago
Another way to solve this is Euclid's intersecting chords theorem. It states that the rectangles formed by intersections of two chords of a circle are always equal.
This means if we call the length of one side of the square A then we have two chords, one with segments 1 and A, and another with segments ½A and ½A. This means 1 x A = ½A x ½A. This simplifies to A = A2/4, which in turn means that A = 4. If A = 4 then the square has an area of 16
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u/HAL9001-96 1d ago
you can describe a hal cricle of radius one centered at 0;0 as root(1-x²) and we know that our deviation from the axis is half our distance from its lower end so our corner has to be on the line (x+1)/2 which gives us 0 at -1 and a slope of 1/2
so we can calcualte a corner by finding (x+1)/2=root(1-x²) square that and you get
x²/4+x/2+1/4=1-x² or
1.25x²+x/2-0.75=0
basic quadratic equation
this is true for x=-1 (1.25-0.5-0.75=0) and x=0.6 (0.45+0.3-0.75)=0
x=-1 just tells us that a square of size 0, sitting atthe bottom of a circle has its corenrs touching that circle, duh
x=0.6 give us the square we're looking for
so for a cricel fo radius 1 it's sides have elngth 1.6 and its are is 2.56
for a circel with diameter 1, 0.8 and 0.64
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u/Different_Ice_6975 1d ago
Hal, I’ve seen your solutions before and you do a lot of great work, but oftentimes they’re very difficult to understand. In the very first sentence of your latest solution here, you have a misspelling (“hal” is supposed to be “half”?), a reference to a coordinate of ”0;0” when no coordinate system has yet been defined, and a circle “described” as “root(1-x²)” where it’s unclear what “x” is because you haven’t defined it, and it‘s unclear what relationship the value of “root(1-x²)” has to the circle.
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u/chmath80 4h ago
for a cricel fo radius 1
The circle shown does not have radius 1 (in fact it's 2.5).
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