1. ## Rectangular

The rectangular WZYX is inside the rectangular ACDB

AB= 8 cm
AC= 6 cm
WX= 8 cm

What is WZ= ?

2. ## Re: 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.

3. ## Re: Rectangular

1 possible solution only:

ax=6.2233
aw=5.027
wz=1.2507
cz=0.78594
wc=0.97296

4. ## Re: Rectangular

Why not (as an attempt) make angle AWX = 60 degrees; then AW = 4 and AX = 4SQRT(3)

5. ## Re: Rectangular

Originally Posted by Mhmh96
The rectangular WZYX is inside the rectangular ACDB

AB= 8 cm
AC= 6 cm
WX= 8 cm

What is WZ= ?
If and only if the points W, X, Y and Z has to be placed on the sides of the rectangle ABCD and $|\overline{WX}| = 8$ Then there is only one unique solution of this question. See attachment #2 to see what happens if $|\overline{AX}|$ is changing but $|\overline{WX}|$ 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:

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

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

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

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

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:

$y \approx 1.895734320$ and consequently
$x \approx 1.133049932$

Since $|\overline{XY}| = |\overline{WZ}|$ you'll get $|\overline{WZ}| \approx \sqrt{1.895734320^2 + 1.133049932^2} \approx 2.208531358$

8. ## Re: Rectangular

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