imc

There are lots of similar triangles in the diagram, but we will only use $△WBZ\sim △ABC.$ If $h$ is the altitude from $B$ to $AC$ and $s$ is the sidelength of the square, then $h-s$ is the altitude from $B$ to $WZ.$ By similar triangles, $$\begin{array}{rl}\frac{h-s}{h}& =\frac{s}{5}\\ 5(h-s)& =hs\\ 5h-5s& =hs\\ 5h& =s(h+5)\\ s& =\frac{5h}{h+5}\end{array}$$ Find the length of the altitude of $△ABC.$ Since it is a right triangle, the area of $△ABC$ is $\frac{1}{2}(3)(4)=6.$ The area can also be expressed as $\frac{1}{2}(5)(h),$ so $\frac{5}{2}h=6⟶h=\mathrm{2.4.}$ Substitute back into $s.$ $$s=\frac{5h}{h+5}=\frac{12}{7.4}=\boxed{\frac{60}{37}}$$