In the diagram, $P R$ and $Q S$ intersect at $V$. Also, $W$ is on $P V, U$ is on $P S$ and $T$ is on $P Q$ with $Q U$ and $S T$ passing through $W$. For some real number $x$,

- the area of $\triangle P U W$ equals $4 x+4$,
- the area of $\triangle S U W$ equals $2 x+20$,
- the area of $\triangle S V W$ equals $5 x+20$,
- the area of $\triangle S V R$ equals $5 x+11$,
- the area of $\triangle Q V R$ equals $8 x+32$, and

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-029.jpg?height=412&width=417&top_left_y=1840&top_left_x=1296)

- the area of $\triangle Q V W$ equals $8 x+50$.

The area of $\triangle P T W$ is closest to
(A) 35
(B) 34
(C) 33
(D) 32
(E) 31
In $\triangle A B C$, if $D$ is on $B C$, then

$$
\frac{\text { Area of } \triangle A B D}{\text { Area of } \triangle A C D}=\frac{B D}{C D}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-052.jpg?height=318&width=420&top_left_y=833&top_left_x=907)

This is because $\triangle A B D$ and $\triangle A C D$ have a common height of $h$ and so

$$
\frac{\text { Area of } \triangle A B D}{\text { Area of } \triangle A C D}==\frac{\frac{1}{2} \times B D \times h}{\frac{1}{2} \times C D \times h}=\frac{B D}{C D}
$$

Using this in the given diagram with point $V$ lying on side $Q S$ of $\triangle W Q S$ and on side $Q S$ of $\triangle R Q S$, we see that

$$
\frac{\text { Area of } \triangle Q V W}{\text { Area of } \triangle S V W}=\frac{Q V}{S V}=\frac{\text { Area of } \triangle Q V R}{\text { Area of } \triangle S V R}
$$

Combining the first and third parts of this equality, we obtain the equivalent equations

$$
\begin{aligned}
\frac{8 x+50}{5 x+20} & =\frac{8 x+32}{5 x+11} \\
(8 x+50)(5 x+11) & =(8 x+32)(5 x+20) \\
40 x^{2}+88 x+250 x+550 & =40 x^{2}+160 x+160 x+640 \\
338 x+550 & =320 x+640 \\
18 x & =90 \\
x & =5
\end{aligned}
$$

Since $x=5$, then we can calculate and fill in the area of most of the pieces of the diagram:

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-052.jpg?height=393&width=415&top_left_y=2210&top_left_x=909)

Let the area of $\triangle P T W$ equal $y$ and the area of $\triangle Q T W$ equal $z$.

We know that

$$
\frac{\text { Area of } \triangle Q V W}{\text { Area of } \triangle S V W}=\frac{Q V}{S V}
$$

and so $\frac{Q V}{S V}=\frac{90}{45}=2$.

Since $V$ lies on side $Q S$ of $\triangle P Q S$, then

$$
\frac{\text { Area of } \triangle P Q V}{\text { Area of } \triangle P S V}=\frac{Q V}{S V}=2
$$

and so $\frac{y+z+90}{99}=2$ which gives $y+z+90=198$ and so $y+z=108$.

Finally, looking at points $W$ on side $T S$ of $\triangle P T S$ and so side $T S$ of $\triangle Q T S$, we get

$$
\frac{y}{54}=\frac{T W}{W S}=\frac{z}{135}
$$

and so $135 y=54 z$ or $5 y=2 z$ or $z=\frac{5}{2} y$.

Therefore, $y+\frac{5}{2} y=108$ or $\frac{7}{2} y=108$ and so $y=\frac{216}{7}=30 \frac{6}{7}$.

Of the given choices, the area of $\triangle P T W$ is closest to 31 .

ANsWer: (E)