+45 In rectangle $WXYZ$, $A$ is on side $\overline{WX}$, $B$ is on side $\overline{YZ}$, and $C$ is on side $\overline{XY}$. If $AX = 2$, $BY = 24$, $\angle ACB= 90^\circ$, and $CY = 3 \cdot CX$, then find $AB$.
+129847 W A 2 X
b
C
3b
Z B 24 Y
AC^2 = b^2 + 2^2 = b^2 + 4
BC^2 = 24^2 + 9b^2 = 9b^2 + 576
AB^2 = AC^2 + BC^2 = 10b^2 + 580
Also
AB^2 = (24 -2)^2 + (b + 3b)^2 = 484 +16b^2
So
16b^2 + 484 = 10b^2 + 580
6b^2 = 96
b^2 = (96/6) = 16
So
AB^2 = 10 (16) + 580
AB^2 = 160 + 580
AB^2 = 740
AB = sqrt [740] = 2sqrt (185) ≈ 27.2
