waffles

Right \(\triangle{ABC}\) has AB = 3, BC = 4, and AC = 5.
Square XYZW is inscribed in triangle ABC with X and Y on \(\overline{AC}\),
W on \(\overline{AB}\), and
Z on \(\overline{BC}\).
What is the side length of the square?

Let s is the side length oft the square \( = \overline{XY} = \overline{YZ} = \overline{ZW} = \overline{WX}\)

Let h = \(\overline{BT}\)

Let A the area of \(\triangle{ABC}\)

h = ?

\(\begin{array}{|rcll|} \hline A &=& \frac{\overline{AB} \cdot \overline{BC} }{2} \\ A &=& \frac{3\cdot 4}{2} \\ \mathbf{A} &\mathbf{=}& \mathbf{6} \\\\ A &=& \frac{\overline{AC}\cdot h}{2} \\ A &=& \frac{5\cdot h}{2} \quad & | \quad \mathbf{A=6} \\ 6 &=& \frac{5\cdot h}{2} \\ \mathbf{h} &\mathbf{=}& \mathbf{ \frac{12}{5} } \\ \hline \end{array}\)

\(\mathbf{\overline{BW} =\ ?}\)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BW} } {s} &=& \frac{ \overline{AB} } { \overline{AC} } \\ \frac{ \overline{BW} } {s} &=& \frac{ 3 } { 5 } \\ \mathbf{ \overline{BW} } & \mathbf{=} & \mathbf{ \frac{3}{5}s } \\ \hline \end{array}\)

\(\mathbf{\overline{BZ} =\ ?} \)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BZ} } {s} &=& \frac{ \overline{BC} } { \overline{AC} } \\ \frac{ \overline{BZ} } {s} &=& \frac{ 4 } { 5 } \\ \mathbf{\overline{BZ}} &\mathbf{=}& \mathbf{\frac{4}{5}s } \\ \hline \end{array}\)

s = ?

\(\begin{array}{|rcll|} \hline A_{\triangle{ZBW}} = \frac{ \overline{BW}\cdot \overline{BZ} } {2} &=& \frac{s\cdot(h-s)} {2} \\ \overline{BW}\cdot \overline{BZ} &=& s\cdot(h-s) \quad & | \quad \mathbf{ \overline{BW} =\frac{3}{5}s } \quad \mathbf{ \overline{BZ} =\frac{4}{5}s } \quad \mathbf{h=\frac{12}{5}} \\ \frac{3}{5}s \cdot \frac{4}{5}s &=& s\cdot(\frac{12}{5}-s) \\ \frac{12}{25}s &=& \frac{12}{5}-s \\ s+\frac{12}{25}s &=& \frac{12}{5} \\ s \cdot \left(1+\frac{12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{25+12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{37}{25} \right) &=& \frac{12}{5} \\ s &=& \frac{25}{37} \cdot \frac{12}{5} \\ s &=& \frac{5}{37} \cdot 12 \\ s &=& \frac{60}{37} \\ \mathbf{s} &\mathbf{=}& \mathbf{1.\overline{621}} \\ \hline \end{array}\)

The side length oft the square is \(\mathbf{1.\overline{621}}\)

Thanks, heureka....here's another way using similar triangles...

Referring to heureka's pic....let the side of the square = s ...let WA  = x  and let BW  = 3 - x

Now....triangles ABC, AXW and WBZ are all similar

Using triangles ABC and AXW, we have that  XW / AW  = BC/ AC →  s / x  =  4 / 5  →  s  = (4/5)x  →  s  = .8x

And using triangles ABC  and WBZ, we have that  BW / ZW  = BA / CA  →  (3 - x ) / s  = 3 / 5  →  (3 - x) / (.8x)  = .6

Multiply both sides by  .8x  and we have that

3 - x  = .48x      add x to  both sides

3  = 1.48x       divide both sides by 1.48

3 / 1.48  = x   =  75 / 37    ..... and s  is .8   of  this  =  600 / 370  = 60/37 units