please help me on this:

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.

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- Jan 6th 2011, 04:39 AMrcsFind the length of the fold. please
please help me on this:

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, 05:51 AMBAdhichange in the diagram
find the value of AB,

- Jan 6th 2011, 06:04 AMSudharaka
- Jan 8th 2011, 06:58 PMrcs
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, 07:07 PMrtblue
Take a look at this rcs:

x+xcosy=4, xsiny=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, 07:54 PMSudharaka
- Jan 8th 2011, 08:08 PMrcs
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, 08:22 PMWilmerCode:
`E 6-b B`

4-a

4

D

a

F 3-b A 3 C

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

AD = 4-a = 25/8

ED = SQRT[(25/6)^2 + (25/8)^2] = 125/24 : 5.208333333...... - Jan 12th 2011, 12:35 AMBAdhi
My answer is (AB/2)Sec(theta)cosec(theta). AB and theta is as given my diagram

- Jan 12th 2011, 09:34 AMWilmer
Ok...but no need to use trig functions: triangles (per my diagram) AEF and ACD are SIMILAR.