Finding the length of line

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June 8th 2012, 02:28 PM
Mhmh96

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Finding the length of line

If you know that AC=3,AB=4,XY=1,then what is XW=?

WXYZ and ABCD are rectangulars
http://store3.up-00.com/June12/zd594451.jpg
June 8th 2012, 09:39 PM
Soroban

Re: Finding the length of line

Hello, Mhmh96!

Quote:

$\text{Given: }\:AC=3,\;AB=4,\;XY=1.$
. . . . . . $W\!XYZ\text{ and }ABDC\text{ are rectangles.}$

$\text{Find the length of }XW.$

http://store3.up-00.com/June12/zd594451.jpg

Draw diagonal $BC.$
Triangle $ABC$ is a 3-4-5 right triangle.

We find that $XW$ and $Y\!Z$ are parallel to $BC.$
And all the right triangles are similar.

$\text{In right triangle }X\!BY\!:\:XY = 1,\;B\!X = \tfrac{3}{5},\;BY = \tfrac{4}{5}$

Can you finish the problem?
June 8th 2012, 09:45 PM
Mhmh96

Re: Finding the length of line

Actually no Soroban,can you explain more .
June 8th 2012, 09:50 PM
Prove It

Re: Finding the length of line

Quote:

Originally Posted by Soroban

Hello, Mhmh96!

Draw diagonal $BC.$
Triangle $ABC$ is a 3-4-5 right triangle.

We find that $XW$ and $Y\!Z$ are parallel to $BC.$
And all the right triangles are similar.

$\text{In right triangle }X\!BY\!:\:XY = 1,\;B\!X = \tfrac{3}{5},\;BY = \tfrac{4}{5}$

Can you finish the problem?

XW and YZ are not necessarily parallel to BC.
June 8th 2012, 10:08 PM
Prove It

Re: Finding the length of line

Quote:

Originally Posted by Mhmh96

If you know that AC=3,AB=4,XY=1,then what is XW=?

WXYZ and ABCD are rectangulars
http://store3.up-00.com/June12/zd594451.jpg

I'm going to call length $\displaystyle \begin{align*} CZ = XB = x \end{align*}$ , length $\displaystyle \begin{align*} WC = BY = y \end{align*}$ and $\displaystyle \begin{align*} WX = YZ = z \end{align*}$ .

From the smallest triangles, we can see that $\displaystyle \begin{align*} x^2 + y^2 = 1 \end{align*}$ by Pythagoras.

We can also see from the larger triangles that

$\displaystyle \begin{align*} (4 - x)^2 + (3 - y)^2 &= z^2 \\ 16 - 8x + x^2 + 9 - 6y + y^2 &= z^2 \\ 16 - 8x + 9 - 6y + 1 &= z^2 \textrm{ since }x^2 + y^2 = 1 \\ 26 - 8x - 6y &= z^2 \\ z &= \sqrt{26 - 8x - 6y} \end{align*}$

Now note that the area of the largest rectangle is $\displaystyle \begin{align*} 12 \end{align*}$ . If we evaluate the area of each smaller piece, we find

$\displaystyle \begin{align*} (3 - y)(4 -x ) + \frac{1}{2}xy + (3 - y)(4 - x) + \frac{1}{2}xy + \sqrt{26 - 8x - 6y} &= 12 \\ 2\left(12 - 3x - 4y + xy\right) + xy + \sqrt{26 - 8x - 6y} &= 12 \\ 24 - 6x - 8y + 2xy + xy + \sqrt{26 - 8x - 6y} &= 12 \\ 24 - 6x - 8y + 3xy + \sqrt{26 - 8x - 6y} &= 12 \end{align*}$

Substituting $\displaystyle \begin{align*} x = \sqrt{1 - y^2} \end{align*}$ , (since we know $\displaystyle \begin{align*} x^2 + y^2 = 1 \end{align*}$ ) will enable solving for $\displaystyle \begin{align*} y \end{align*}$ , which will enable solving for $\displaystyle \begin{align*} x \end{align*}$ , and finally solving for $\displaystyle \begin{align*} z \end{align*}$ .

See how you go.
June 10th 2012, 03:40 AM
bjhopper

Re: Finding the length of line

Following Soroban's hint I calculated

WX= 4.25= ( 17/5) / cos 36.87
ZY = 3.66 = (11/5 /sin 36.87

WXYZ is not a rectangle but a trapezoid
June 10th 2012, 05:42 AM
BobP

Re: Finding the length of line

The rectangle has to be in a particular position, the angle AXW being approximately 31.7501 degrees.
WX turns out to be approximately 4.08512.