In triangle $ABC$, we have that $E$ and $F$ are midpoints of sides $\overline{AC}$ and $\overline{AB}$, respectively. The area of $\triangle ABC$ is 24 square units. How many square units are in the area of $\triangle CEF$?
Call AB the height of ΔABC and AC the base
Then (1/2) AC * AB = 24 units^2 ⇒ AC * AB = 48 units^2
But EC = (1/2)AC is the base of ΔCEF and (1/2)AB = AF is the height
So area of Δ CEF = (1/2)CE *AF = (1/2) [(1/2)AC] * [(1/2)AB] = 1/8 AC * AB =
(1/8)* 48 = 6 units^2
+618 
