## Math Apps: A square piece of paper ABCD is black on one side, white on the other side

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AerialAtom
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### Math Apps: A square piece of paper ABCD is black on one side, white on the other side

So for my final math assignment the teacher assigned a Math application sheet and I'm stumped on this problem.

A square piece of paper ABCD is black on one side and white on the other side and has an area of 3 in^2.

Corner A is folded over to point E which lies on the diagonal AC such that the total visible area is half black and half white.

How far is E from the fold line?

anonmeans
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### Re: Math Applications

A square piece of paper ABCD is black on one side and white on the other side and has an area of 3 in^2.

Corner A is folded over to point E which lies on the diagonal AC such that the total visible area is half black and half white.
So you're starting with this:

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```B*---------*C | / | | / | | / | | / | A*---------*D```
And you're folding to get this;

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```B*---------*C | E / | |____*/ | \ | | \ | | -----*D```
(Please let me know if this isn't right.)
How far is E from the fold line?
If the areas are equal then you have to find the area of the fold-over triangle so it matches the left-over of the square. Label the ends of the fold lines so we can describe things easier:

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```B*---------*C | E / | |____*/ | M \ | | \ | | N-----*D ```
Unfold this to see this:

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```B*---------*C | E / | |____*/ | M|\ | | | \ | | A*---N-----*D```
MEN = BCDNEM and MEN = MAN. The area of the first square is BCDNEM + MEN + MAN = 3*MAN = 3 sq in. MAN is 45-45-90 triangle so it has the ratios they show here. The area is (1/2)(AN)(AM) = 3 = (1/2)(b)(h). But b = h so 3 = (1/2)(h^2). So what is h? The fold line is MN. E is half this length because MN = AE and MANE is a square. So what is this distance?

Please write back if you get stuck. Thanks.

AerialAtom
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Joined: Sat Dec 05, 2015 2:26 am
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### Re: Math Apps: A square piece of paper ABCD is black on one side, white on the other side

So I'm going to take this 3 = (1/2)(h^2) and solve

$3 = (1/2)(h^2) 3*2=2*(1/2)(h^2) 6=h^2 h=sqrt(6)$

since we got h the we can put that into the pythagorean theorem, right?

$a^2+b^2=c^2 (sqrt(6))^2+(sqrt(6))^2=c^2 6+6=c^2 12=c^2 sqrt(12)=c sqrt(4*3)=c 2sqrt(3)=c$

and since
$MN=AE AE=2sqrt3$

Right?