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# This is a very hard problem, lets see if ANYONE (Even Moderators) Can try to figure this out.

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What is the area of the red?

Nov 19, 2017

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Both Circles are the same size, the box is made of perfect 90° angles,

Here is an update, left something out

Nov 19, 2017
edited by ProMagma  Nov 19, 2017
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I think there is still something missing because you have not told us how big the overlap is.

I think to get a number answer we would need that, or something that would allow us to find it..

I suppose I can assume there is a 1.5inch border around the circles which would mean that the circles are tight inside a rectangle  12 inches high and 17inches long

Nov 19, 2017
edited by Melody  Nov 19, 2017
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Hi ProMagma,

I have had to make some assumptions here. They are stated in my last post.

I have drawn this to scale.

$$cos\theta=\frac{2.5}{6}=\frac{5}{12}\\ \theta \approx 65.376^\circ\\ \therefore \angle AOB\approx 130.751^\circ$$

$$Area\;\;of\;\; \triangle AOB\approx 0.5*6*6*sin130.751^\circ\approx 13.636 u^2\\ Area\;\;of\;\; sector \;\;AOB\approx \frac{130.751}{360}\pi *6^2 \approx 41.077u^2\\ Area\;\;of\;\; minor \;\; segment \;\;AB\approx 41.077-13.636 = 27.441u^2$$

This is the same as the Red region inside the intersection of the circles because both are half of the total intersection.

Now the area of the union of the 2 circles is   $$2\pi r^2 - 2*27.441 \approx 226.195 - 54.882\approx 171.313$$

Half of this area (because the rectangle is split in half) is $$\approx 85.857$$

The area of the whole rectangle is  $$20*15=300 inch^2$$

The are of half the rectangle is  $$150inch^2$$

So

$$Original\;red\;shaded\:area\;\;\approx 150 - 85.857 + 27.441 \approx 91.584 \;inch^2$$

I think that is right but my logic, and arithemetic does need to be double checked.

Nov 19, 2017
edited by Melody  Nov 19, 2017
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This appears to be a variation of the problem posed here: "Mind Your Decisions- Two Circles within a Rectangle.", on YouTube.

Nov 19, 2017
edited by Guest  Nov 19, 2017
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Thanks guest but did you forget to include the link?

Melody  Nov 20, 2017
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Sorry Melody: I did forget to give the link. Here it is:

Nov 20, 2017
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Thanks :)

Melody  Nov 20, 2017
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Yes, but it is not the exact math problem

ProMagma  Nov 21, 2017
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Hi ProMagna,

Well we cannot give you the exact answer if you do not give us the exact question...

OR were you referring to the you tube link?

Did you understand my explanation?

Melody  Nov 21, 2017
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Yes,  That was very good,

I did understand the explination,

thank you Melody

ProMagma  Nov 21, 2017
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And yes, I was reffering to the tube link

ProMagma  Nov 21, 2017