Let $t$ be TNYWR.

In the diagram, point $C$ is on $B D, \triangle A B C$ is right-angled at $B, \triangle A C E$ is right-angled at $C$, and $\triangle C D E$ is right-angled at $D$. Also, $A B=2 t, B D=D E=9 t$, and $B C: C D=2: 1$. If the area of $\triangle A C E$ is $k$, what is the value of $\frac{1}{36} k$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_23a04f66c1167e793e90g-3.jpg?height=466&width=477&top_left_y=1187&top_left_x=846)
Since $B C: C D=2: 1$, then $B C=2 C D$.

Since $B D=9 t$, then $B C+C D=9 t$ or $C D+2 C D=9 t$ and so $3 C D=9 t$ or $C D=3 t$, which gives $B C=6 t$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-18.jpg?height=474&width=498&top_left_y=2148&top_left_x=911)

The area of $\triangle A C E$, which is denoted $k$, equals the area of trapezoid $A B D E$ minus the areas of $\triangle A B C$ and $\triangle C D E$.

The area of trapezoid $A B D E$ is $\frac{1}{2}(A B+D E)(B D)$ or $\frac{1}{2}(2 t+9 t)(9 t)$ which equals $\frac{99}{2} t^{2}$.

The area of $\triangle A B C$ is $\frac{1}{2}(A B)(B C)$ or $\frac{1}{2}(2 t)(6 t)$ which equals $6 t^{2}$.

The area of $\triangle C D E$ is $\frac{1}{2}(C D)(D E)$ or $\frac{1}{2}(3 t)(9 t)$ which equals $\frac{27}{2} t^{2}$.

Therefore, $k=\frac{99}{2} t^{2}-6 t^{2}-\frac{27}{2} t^{2}=30 t^{2}$.

Since the answer to (a) is 6 , then $t=6$, and so $\frac{1}{36} k=\frac{30}{36} t^{2}=30$.