Find the length of the fold. please

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January 6th 2011, 03:39 AM
rcs

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Find 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.
January 6th 2011, 04:51 AM
BAdhi

1 Attachment(s)

change in the diagram

find the value of AB, $\theta$
January 6th 2011, 05:04 AM
Sudharaka

1 Attachment(s)

Quote:

Originally Posted by rcs

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.

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.
January 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?
January 8th 2011, 06:07 PM
rtblue

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.
January 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?

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

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

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

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

$\theta=36.87^o$
January 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...
January 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
AD = 4-a = 25/8
ED = SQRT[(25/6)^2 + (25/8)^2] = 125/24 : 5.208333333......
January 11th 2011, 11:35 PM
BAdhi

My answer is (AB/2)Sec(theta)cosec(theta). AB and theta is as given my diagram
January 12th 2011, 08:34 AM
Wilmer

Ok...but no need to use trig functions: triangles (per my diagram) AEF and ACD are SIMILAR.