The area of rectangle $ABCD$ is 72. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, what is the area of the triangle?
Three right triangles lie outside $\triangle AMN$. Their areas are $\frac{1}{4}$, $\frac{1}{4}$, and $\frac{1}{8}$ for a total of $\frac{5}{8}$ of the rectangle. The area of $\triangle AMN$ is $\frac{3}{8}(72)=27$.

OR

Let the rectangle have sides of $2a$ and $2b$ so that $4ab=72$ and $ab=18$.Three right triangles lie outside triangle $AMN$, and their areas are $\frac{1}{2}(2a)(b)$, $\frac{1}{2}(2b)(a)$, $\frac{1}{2}(a)(b)$, for a total of $\frac{5}{2}(ab)=\frac{5}{2}(18)=45$. The area of triangle $AMN$ is $72-45=27$.

[asy]
/* AMC8 2000 #25 Solution */
pair A=(0,1), B=(1.5,1), C=(1.5,0), D=(0,0);
draw(A--B--C--D--cycle);
draw((.75,0)--(0,1)--(1.5,.5)--cycle);
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
label("$N$", (0.75,0), S, red);
label("$M$", (1.5,.5), E, red);
[/asy]