$\triangle P Q R$ has $P Q=150$ and $P R=Q R=125$, as shown. Three line segments are drawn parallel to $Q R$, dividing $\triangle P Q R$ into four sections of equal area. The height, $h$, of the bottom section is closest to
(A) 16.7
(B) 16.9
(C) 16.5
(D) 16.3
(E) 16.1

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-059.jpg?height=524&width=487&top_left_y=1077&top_left_x=1231)
We label the other two vertices of the bottom section as $S$ (on $P R$ ) and $T$ (on $P Q$ ).

First, we calculate the area of $\triangle P Q R$.

We do this by dropping a perpendicular from $R$ to $P Q$.

Since $P R=Q R$, then $\triangle P Q R$ is isosceles and the perpendicular from $R$ to $P Q$ meets $P Q$ at its midpoint $M$.

Thus, $P M=M Q=\frac{1}{2}(150)=75$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-117.jpg?height=518&width=507&top_left_y=451&top_left_x=863)

By the Pythagorean Theorem,

$$
R M^{2}=R Q^{2}-M Q^{2}=125^{2}-75^{2}=15625-5625=10000
$$

Since $R M>0$, then $R M=\sqrt{10000}=100$.

Therefore, the area of $\triangle P Q R$ is $\frac{1}{2}(R M)(P Q)=\frac{1}{2}(100)(150)=7500$.

Let $F$ be the point on $Q R$ for which $P F$ is perpendicular to $Q R$ and let $G$ be the point where $P F$ intersects $T S$ (also at a $90^{\circ}$ angle).

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-117.jpg?height=522&width=511&top_left_y=1341&top_left_x=861)

Then the area of $\triangle P Q R$ is also equal to $\frac{1}{2}(Q R)(P F)$.

Thus, $\frac{1}{2}(125)(P F)=7500$ and so $P F=\frac{15000}{125}=120$.

Since $T S$ is parallel to $Q R$, then $\angle P T S=\angle P Q R$ and $\angle P S T=\angle P R Q$.

This means that $\triangle P T S$ is similar to $\triangle P Q R$.

Since each of the four sections of $\triangle P Q R$ is one-quarter of the total area, then $\triangle P T S$ is threequarters of the total area of $\triangle P Q R$.

This means that the dimensions of $\triangle P T S$ are $\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$ of those of $\triangle P Q R$. (In general, if two similar triangles have areas in the ratio $k: 1$, then their corresponding sides are in the ratio $\sqrt{k}: 1$.)

Therefore, $P G=\frac{\sqrt{3}}{2} P F=\frac{\sqrt{3}}{2}(120)=60 \sqrt{3}$.

Finally, since $T S$ is parallel to $Q R$, then $h=G F=P F-P G=120-60 \sqrt{3} \approx 16.077$.

Of the given answers, the height is closest to 16.1.