A Rectangular Piece of Paper is folded

**D. Martin** · June 1st 2008, 02:04 PM

The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.

10 X 6
B________________________ A
| / |
| / |
| / |
| ________/______________ |
C Y D

I realize that both AB and CD are 16 across. I've tried turning parts of this rectangel into a triangle. But that didn't yield results, as I couldn't determine the new figure's length. The closest I got to a lead was My thought of drawing a line from Y to A. That would give me a triangle with a base of 6, but no angles, so no way to determine XY.

**D. Martin** · June 1st 2008, 02:14 PM

I just added the diagram. I couldn't make it on this form.

**Jonboy** · June 1st 2008, 04:45 PM

Since you have a rectangle, the length is half the width. So $BC$ and $AD$ are 8.

And you can make this picture: link

Now apply the Pythagorean Theorem a bunch of times.

$(XB)^2 + (BC)^2 = (XC)^2$

$8^2 + 10^2 = (XC)^2$

$164 = (XC)^2$

$XC = 2\sqrt{41}$

Then apply it again: $(XA)^2 + (AD)^2 = (DX)^2$

$6^2 + 8^2 = (DX)^2$

$36 + 64 = (DX)^2$

$80 = (DX)^2$

$DX = 4\sqrt{5}$

Now let's fill that in our pic: link

Since $XZ$ is an angle bisector, then $ZC = DZ$

So both are 8.

Apply Pythagorean Theorem to get $XZ$

$(ZC)^2 + (XZ)^2 = (2\sqrt{41})^2$

$64 + (XZ)^2 = 164$

$(XY)^2 = 100$

$XY = 10$

And without more info, I don't see how you can find $XY$

**D. Martin** · June 1st 2008, 07:16 PM

Thank you, very much. That was really helpful and I wouldn't have figured out the problem by myself before tomorrow. So I really appreciate your assistance.

**D. Martin** · June 1st 2008, 07:28 PM

Originally Posted by Jonboy

Since you have a rectangle, the length is half the width. So $BC$ and $AD$ are 8.

And you can make this picture: link

Now apply the Pythagorean Theorem a bunch of times.

$(XB)^2 + (BC)^2 = (XC)^2$

$8^2 + 10^2 = (XC)^2$

$164 = (XC)^2$

$XC = 2\sqrt{41}$

Then apply it again: $(XA)^2 + (AD)^2 = (DX)^2$

$6^2 + 8^2 = (DX)^2$

$36 + 64 = (DX)^2$

$80 = (DX)^2$

$DX = 4\sqrt{5}$

Now let's fill that in our pic: link

Since $XZ$ is an angle bisector, then $ZC = DZ$

So both are 8.

Apply Pythagorean Theorem to get $XZ$

$(ZC)^2 + (XZ)^2 = (2\sqrt{41})^2$

$64 + (XZ)^2 = 164$

$(XY)^2 = 100$

$XY = 10$

And without more info, I don't see how you can find $XY$

I'm just wondering, how did you deduce that the length was half the width?

**Jonboy** · June 2nd 2008, 04:36 AM

Honestly, I assumed too much.

The only way I could've got the width was knowing the perimeter, which we don't know.

That's because:

$P = 2L + 2W$

$P = 2(16) + 2W$

$2W = P - 32$

$W = \frac{P - 32}{2} = \frac{1}{2}P - 16$

But, look on the bright side, it further proves you can't solve this problem.

**topsquark** · June 2nd 2008, 04:44 AM

Originally Posted by D. Martin

The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.

10 X 6
B________________________ A
| / |
| / |
| / |
| ________/______________ |
C Y D

I realize that both AB and CD are 16 across. I've tried turning parts of this rectangel into a triangle. But that didn't yield results, as I couldn't determine the new figure's length. The closest I got to a lead was My thought of drawing a line from Y to A. That would give me a triangle with a base of 6, but no angles, so no way to determine XY.

I don't have time to get to it right now, but I'll try (no promises) to get back to it later.

Note that whereas we don't know the width of the rectangle, we do know that points B and D lie the same distance away from the fold along the red line (the diagonal) because when the paper is folded these points overlap. My diagram stinks, but it gives you the idea behind what I'm talking about anyway.

Edit: Oh! I almost forgot. BD and XY have to be perpendicular as well. Do you see why that is?

-Dan

**earboth** · June 2nd 2008, 10:33 AM

Originally Posted by D. Martin

The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.

10 X 6
B________________________ A
| / |
| / |
| / |
| ________/______________ |
C Y D

I've sketched the rectangle.

BX = XD = 10 and XA = 6. Use Pythagorean theorem in the triangle XAD. Therefore AD = 8.

Use Pythagorean theorem with the grey triangle to calculate the length of the crease. I've got $c = \sqrt{16+64} = 4 \cdot \sqrt{5}$

**goldbh** · June 7th 2008, 01:25 PM

I'd forgotten how much I love this stuff.

Thread: A Rectangular Piece of Paper is folded

LinkBack

Thread Tools

Display

A Rectangular Piece of Paper is folded

Thank you

Similar Math Help Forum Discussions

Calculus word problem: folded paper

For a piece of code

find g(X) that transforms folded normal into gamma(alpha,beta)

A Piece of Wire

Portfolio Piece Question

Search Tags