$\overline {BE}$ is the midsegment of $\Delta ACD$, shown at right.  

1. Find the perimeter of $\Delta ACD$.

   Hint (a):

   The perimeter of $\Delta ACD = (2)(\text{ the perimeter of } \Delta ABE)$.
   Since BE is a midsegment, the two triangle have a zoom factor of $2$, that is why it is multiplied by $2$.

2. If the area of $\Delta ABE$ is $19 \; \text{cm}^2$, what is the area of $\Delta ACD$?

   Hint (b):

   The area of $\Delta ACD = (\text{the area of } \Delta ABE)(\text { the zoom factor)}^²$.