In rectangle $WXYZ$, $A$ is on side $\overline{WX}$, $B$ is on side $\overline{YZ}$, and $C$ is on side $\overline{XY}$. If $AX = 2$, $CX = 24$, $CY = 3$, $BY = 3$, then find the area of triangle $ABC$.
W A 2 X
24
C
3
Z B 3 Y
AXYB is a trapezoid
Area = (1/2) (XC + CY) (AX + BY) = (1/2) (24 + 3) ( 2 + 3) = 135/2 = 67.5
Triangle AXC has area (1/2((2* 24) = 12
Triangle CYB has area (1/2) (3 * 3) = 4.5
[ABC ] = [ AYXB ] - [ AXC ] - [ CYB ] = 67.5 - 12 - 4.5 = 51
+1431 
