11. The area of triangle $A B C$ is $60 \mathrm{~m}^{2}$. Let $D, E$ be the points on the side $A B$ such that $A D=D E=E B$ and let $F, G, H$ be the points on $A C$ such that $A F=F G=G H=H C$. What is the area of triangle $A E H$?
(A) $20 \mathrm{~m}^{2}$
(B) $45 \mathrm{~m}^{2}$
(C) $30 \mathrm{~m}^{2}$
(D) $40 \mathrm{~m}^{2}$
(E) $24 \mathrm{~m}^{2}$
(11) The correct answer is $(C)$.

Since $A E$ is equal to $2 / 3$ of $A B$, the area of triangle $A E C$ is $2 / 3$ of that of $A B C$ (indeed, the height from $C$ is the same for both triangles). Similarly, the area of $A E H$ is equal to $3 / 4$ of the area of $A E C$. Therefore, the area of $A E H$ is equal to $\frac{2}{3} \cdot \frac{3}{4}=\frac{1}{2}$ of the area of triangle $A B C$.

Problem proposed by Matteo Protopapa.