1. ## Find the length of the fold. please

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.

2. ## change in the diagram

find the value of AB, $\displaystyle \theta$

3. Originally Posted by rcs

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, $\displaystyle x+x\cos\theta=4~and~x\sin\theta=3$ Solving these you can find x and $\displaystyle \theta$. Hope you would be able to continue from here.

4. 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?

5. 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.

6. 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?
$\displaystyle x+x\cos\theta=4~and~x\sin\theta=3$

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

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

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

$\displaystyle \theta=36.87^o$

7. thank you... i can solve now for the x side. then use other trig function to be able to find the longer side...

8. 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
ED = SQRT[(25/6)^2 + (25/8)^2] = 125/24 : 5.208333333......

9. My answer is (AB/2)Sec(theta)cosec(theta). AB and theta is as given my diagram

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

,

,

### an 8 inch paper is folded so that upper right corner touches the middle of the opposite side.

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