# A Rectangular Piece of Paper is folded

• Jun 1st 2008, 02:04 PM
D. Martin
A Rectangular Piece of Paper is folded
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.
• Jun 1st 2008, 02:14 PM
D. Martin
I just added the diagram. I couldn't make it on this form.
• Jun 1st 2008, 04:45 PM
Jonboy
Since you have a rectangle, the length is half the width. So $\displaystyle BC$ and $\displaystyle AD$ are 8.

And you can make this picture: link

Now apply the Pythagorean Theorem a bunch of times.

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

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

$\displaystyle 164 = (XC)^2$

$\displaystyle XC = 2\sqrt{41}$

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

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

$\displaystyle 36 + 64 = (DX)^2$

$\displaystyle 80 = (DX)^2$

$\displaystyle DX = 4\sqrt{5}$

Now let's fill that in our pic: link

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

So both are 8.

Apply Pythagorean Theorem to get $\displaystyle XZ$

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

$\displaystyle 64 + (XZ)^2 = 164$

$\displaystyle (XY)^2 = 100$

$\displaystyle XY = 10$

And without more info, I don't see how you can find $\displaystyle XY$
• Jun 1st 2008, 07:16 PM
D. Martin
Thank you
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.
• Jun 1st 2008, 07:28 PM
D. Martin
Quote:

Originally Posted by Jonboy
Since you have a rectangle, the length is half the width. So $\displaystyle BC$ and $\displaystyle AD$ are 8.

And you can make this picture: link

Now apply the Pythagorean Theorem a bunch of times.

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

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

$\displaystyle 164 = (XC)^2$

$\displaystyle XC = 2\sqrt{41}$

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

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

$\displaystyle 36 + 64 = (DX)^2$

$\displaystyle 80 = (DX)^2$

$\displaystyle DX = 4\sqrt{5}$

Now let's fill that in our pic: link

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

So both are 8.

Apply Pythagorean Theorem to get $\displaystyle XZ$

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

$\displaystyle 64 + (XZ)^2 = 164$

$\displaystyle (XY)^2 = 100$

$\displaystyle XY = 10$

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

I'm just wondering, how did you deduce that the length was half the width?
• Jun 2nd 2008, 04:36 AM
Jonboy
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:

$\displaystyle P = 2L + 2W$

$\displaystyle P = 2(16) + 2W$

$\displaystyle 2W = P - 32$

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

But, look on the bright side, it further proves you can't solve this problem. ;)
• Jun 2nd 2008, 04:44 AM
topsquark
Quote:

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
• Jun 2nd 2008, 10:33 AM
earboth
Quote:

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 $\displaystyle c = \sqrt{16+64} = 4 \cdot \sqrt{5}$
• Jun 7th 2008, 01:25 PM
goldbh
I'd forgotten how much I love this stuff. (Clapping)