# Find the length of the fold. please

• Jan 6th 2011, 03:39 AM
rcs
Find the length of the fold. please

Attachment 20340

A 4 by 6 inch paper is folded so that its upper right corner touches
the midpoint of an opposite side and such that the fold obtained is the
longer one. Find the length of the fold.
• Jan 6th 2011, 04:51 AM
change in the diagram
find the value of AB, $\theta$
• Jan 6th 2011, 05:04 AM
Sudharaka
Quote:

Originally Posted by rcs

Attachment 20340

A 4 by 6 inch paper is folded so that its upper right corner touches
the midpoint of an opposite side and such that the fold obtained is the
longer one. Find the length of the fold.

Dear rcs,

Look at the figure I had drawn. You can obtain the two equations, $x+x\cos\theta=4~and~x\sin\theta=3$ Solving these you can find x and $\theta$. Hope you would be able to continue from here.
• Jan 8th 2011, 05:58 PM
rcs
how can the theta be solved if it has only one side that has an exact value? x + xsin theta = 4 and xsin theta = 3 ? elimination?
• Jan 8th 2011, 06:07 PM
rtblue
Take a look at this rcs:

x&#43;xcosy&#61;4, xsiny&#61;3 - Wolfram|Alpha

The system of equations can be solved by graphing the individual equations, and looking for an intersection point. There is only one.
• Jan 8th 2011, 06:54 PM
Sudharaka
Quote:

Originally Posted by rcs
how can the theta be solved if it has only one side that has an exact value? x + xsin theta = 4 and xsin theta = 3 ? elimination?

$x+x\cos\theta=4~and~x\sin\theta=3$

$3\displaystyle\frac{(1+\cos\theta)}{\sin\theta}=4$

$3\displaystyle\left(\frac{2\cos^2\frac{\theta}{2}} {2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right) =4$

$\cot\displaystyle\frac{\theta}{2}=\frac{4}{3}$

$\theta=36.87^o$
• Jan 8th 2011, 07:08 PM
rcs
thank you... i can solve now for the x side. then use other trig function to be able to find the longer side...
• Jan 8th 2011, 07:22 PM
Wilmer
Code:

E          6-b              B                             4-a 4                               D                                                             a F 3-b  A        3            C
We only need to call on Big Pete Pythagoras:

My points A and B are same as yours; the "fold line" is ED.
I'm leaving left side of diagram out: making its length b, so EB = EA = 6-b;
plus length of AF becomes 3-b.
Let CD = a; then DB = DA = 4-a

Right triangle AEF: 4^2 + (3-b)^2 = (6-b)^2 : leads to b = 11/6

Right triangle ACD: 3^2 + a^2 = (4-a)^2 : leads to a = 7/8

We can now use right triangle ADE (or BDE) to get fold line ED; using ADE:
AE = 6-b = 25/6