The rectangular WZYX is inside the rectangular ACDB

AB= 8 cm

AC= 6 cm

WX= 8 cm

What is WZ= ?

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- Aug 24th 2012, 02:33 PMMhmh96Rectangular
The rectangular WZYX is inside the rectangular ACDB

AB= 8 cm

AC= 6 cm

WX= 8 cm

What is WZ= ?

http://store2.up-00.com/Aug12/yLN47559.jpg - Aug 24th 2012, 02:50 PMHallsofIvyRe: Rectangular
There are an infinite number of possible solutions. Imagine sliding point X toward A or B. You can aways rotate line XW around X so that W stays on AC. Each angle gives a different solution.

- Aug 24th 2012, 03:31 PMMaxJasperRe: Rectangular
1 possible solution only:

ax=6.2233

aw=5.027

wz=1.2507

cz=0.78594

wc=0.97296 - Aug 25th 2012, 02:55 AMWilmerRe: Rectangular
Why not (as an attempt) make angle AWX = 60 degrees; then AW = 4 and AX = 4SQRT(3)

- Aug 25th 2012, 10:39 AMearbothRe: Rectangular
If and only if the points W, X, Y and Z has to be placed on the sides of the rectangle ABCD

**and**$\displaystyle |\overline{WX}| = 8$ Then there is only one unique solution of this question. See attachment #2 to see what happens if $\displaystyle |\overline{AX}| $ is changing but $\displaystyle |\overline{WX}| $ remains constant. Only the thick outlined quadrilateral is a rectangle, all other quadrilaterals are trapezoids.

According to the labeling in attachment #1 you'll get:

$\displaystyle 8^2 = (8 - x)^2 + (6 - y)^2$ ...........**[1]**

The adjacent sides of the rectangle are perpendicular to each other. Using the slopes of the sides you'll get:

$\displaystyle -\frac yx \cdot \frac{6-y}{8-x}=-1$ ...........**[2]**

From [2] you'll get: $\displaystyle x = 4 - \sqrt{y^2 - 6·y + 16}$

Plug in the term of x into [1] and solve for y. I've to confess that this task was done by my calculator. You'll get:

$\displaystyle y \approx 1.895734320$ and consequently

$\displaystyle x \approx 1.133049932$

Since $\displaystyle |\overline{XY}| = |\overline{WZ}|$ you'll get $\displaystyle |\overline{WZ}| \approx \sqrt{1.895734320^2 + 1.133049932^2} \approx 2.208531358$ - Nov 1st 2012, 01:48 AMelisaevedentRe: Rectangular
- Nov 8th 2012, 07:21 PMChalama123Re: Rectangular
- Nov 18th 2012, 11:35 PMjesongRe: Rectangular
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