3. In the diagram, $A B C D$ is a rectangle with points $P$ and $Q$ on $A D$ so that $A B=A P=P Q=Q D$. Also, point $R$ is on $D C$ with $D R=R C$. If $B C=24$, what is the area of $\triangle P Q R$ ?
3. Solution 1
Since $A B C D$ is a rectangle and $B C=24$, then $A D=24$.
Since $A D$ is divided into three equal parts $A P=P Q=Q D$, then each part has length $24 \div 3=8$.
Thus, $A B=A P=P Q=Q D=8$.
Since $A B C D$ is a rectangle and $A B=8$, then $D C=8$.
Since $R$ is the midpoint of $D C$, then $D R=\frac{1}{2}(8)=4$.
Now $\triangle P Q R$ can be viewed as having base $P Q$. In this case, $D R$ is the height of $\triangle P Q R$, since it is the perpendicular distance from $R$ to the line through $P$ and $Q$.
Therefore, the area of $\triangle P Q R$ is $\frac{1}{2}(P Q)(D R)=\frac{1}{2}(8)(4)=16$.
Solution 2
We begin as in Solution 1 and determine that $A B=A P=P Q=Q D=8$ and $D R=4$. Now, we note that the area of $\triangle P Q R$ equals the area of $\triangle P D R$ minus the area of $\triangle Q D R$. Each of these triangles is right-angled at $D$.
$\triangle P D R$ has $P D=8+8=16$ and $D R=4$, and so has area $\frac{1}{2}(16)(4)=32$.
$\triangle Q D R$ has $Q D=8$ and $D R=4$, and so has area $\frac{1}{2}(8)(4)=16$.
Thus, the area of $\triangle P Q R$ is $32-16=16$.
ANSWER: 16