Right triangle \(\triangle ABC\) has side lengths 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 the side of the square be = S
Because XW = YZ.....then WZ is parallel to AC
Triangle ABC is right with angle ABC = 90°
So....
angle BZW = angle BCA
angle BWZ = angle BAC
And angle ZYC = angle WXA = angle ABC
Triangle YCZ is similar to triangle BCA by AA congruency
So
YC / ZY = BC/BA
YC / S = 4/3
YC = (4/3)S
And triangle XAW is similar to triangle BAC by AA congruency
So
AX / XW = BA / BC
AX / S = 3 / 4
AX = (3/4)S
And
AX + XY + YC = 5
So
(3/4)S + S + (4/3)S = 5
S [ 3/4 + 4/3 + 1 ] = 5
S [ 9/12 + 16/12 + 12/12 ] / 12 = 5
S [ 37/12] = 5
S = 5 (12) /37 = 60 / 37