8. Square $A B C D$ has sides of length $3 \mathrm{~cm}$. The points $M$ and $N$ lie on $A D$ and $A B$ so that $C M$ and $C N$ split the square into three pieces of the same area. What is the length of $D M$ ?
A $0.5 \mathrm{~cm}$
B $1 \mathrm{~cm}$
C $1.5 \mathrm{~cm}$
D $2 \mathrm{~cm}$
E $2.5 \mathrm{~cm}$
8. D The area of square $A B C D$ is $(3 \times 3) \mathrm{cm}^{2}=9 \mathrm{~cm}^{2}$. Hence the area of each piece is $\frac{1}{3} \times 9 \mathrm{~cm}^{2}=3 \mathrm{~cm}^{2}$. Since the area of a triangle is equal to half its base multiplied by its perpendicular height, we have $\frac{1}{2} \times D M \times D C=3 \mathrm{~cm}^{2}$. Therefore $\frac{1}{2} \times D M \times 3 \mathrm{~cm}=3 \mathrm{~cm}^{2}$ and hence the length of $D M$ is $2 \mathrm{~cm}$.