Solving a complex set of triangles, I think.

[travkliewer]

It's been decades since I was in school. And Math was my favorite subject. Anyways, I was just trying to solve a simple problem at work, but now it's a vendetta.
Referring to attached image. We know X and Y, but can we figure out and of the other values? I've tried various triangle solving techniques, but like I said, it's been a long time.

 

[skeeter]

a = d and b = c

Given x and y, you would also need one of a, b, c, d, or z

The attached diagram shows two rectangles with identical width (x) and height (y). Note the parallelograms are not the same size.

 

[travkliewer]

x, y and z are given.

 

[skeeter]

reference the attached diagram ...

Pythagoras yields [imath](x-c)^2+y^2 = a^2[/imath]

the small right triangle with hypotenuse = c and leg = z is similar to either of the two larger, congruent right triangles
similar triangles have sides in proportion [imath]\implies \dfrac{a}{y}=\dfrac{c}{z} \implies a = \dfrac{cy}{z}[/imath]

substituting for [imath]a[/imath] in the Pythagoras equation yields
[imath](x-c)^2 + y^2 = \left(\dfrac{y}{z}\right)^2 \cdot c^2[/imath], an equation quadratic in [imath]x[/imath].

The grunt work algebra using the quadratic formula yields

[imath] c = \dfrac{-xz^2 + yz\sqrt{x^2+y^2-z^2}}{y^2-z^2}[/imath]

of course, [imath]c=b[/imath] and you can find [imath]a=d[/imath]

 

[skeeter]

Edit … equation is quadratic in c