In the diagram, $P Q=Q R=R S=S P=S Q=6$ and $P T=R T=14$. The length of $S T$ is
(A) $4 \sqrt{10}-3$
(B) 11
(C) $7 \sqrt{3}-3$
(D) 10
(E) $\sqrt{232-84 \sqrt{3}}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-082.jpg?height=311&width=458&top_left_y=598&top_left_x=1294)
Solution 1

First, we note that $\triangle P Q S$ and $\triangle R Q S$ are equilateral.

Join $P$ to $R$. Since $P Q R S$ is a rhombus, then $P R$ and $Q S$ bisect each other at their point of intersection, $M$, and are perpendicular.

Note that $Q M=M S=\frac{1}{2} Q S=3$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-162.jpg?height=309&width=461&top_left_y=165&top_left_x=886)

Since $\angle P S Q=60^{\circ}$, then $P M=P S \sin (\angle P S M)=6 \sin \left(60^{\circ}\right)=6\left(\frac{\sqrt{3}}{2}\right)=3 \sqrt{3}$.

Since $P T=T R$, then $\triangle P R T$ is isosceles.

Since $M$ is the midpoint of $P R$, then $T M$ is perpendicular to $P R$.

Since $S M$ is also perpendicular to $P R$, then $S$ lies on $T M$.

By the Pythagorean Theorem in $\triangle P M T$, since $M T>0$, we have

$$
M T=\sqrt{P T^{2}-P M^{2}}=\sqrt{14^{2}-(3 \sqrt{3})^{2}}=\sqrt{196-27}=\sqrt{169}=13
$$

Therefore, $S T=M T-M S=13-3=10$.

## Solution 2

First, we note that $\triangle P Q S$ and $\triangle R Q S$ are equilateral.

Join $P$ to $R$. Since $P Q R S$ is a rhombus, then $P R$ and $Q S$ bisect each other at their point of intersection, $M$, and are perpendicular.

Since $P T=T R$, then $\triangle P R T$ is isosceles.

Since $M$ is the midpoint of $P R$, then $T M$ is perpendicular to $P R$.

Since $S M$ is also perpendicular to $P R$, then $S$ lies on $T M$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-162.jpg?height=309&width=458&top_left_y=1434&top_left_x=888)

Since $\angle P S Q=60^{\circ}$, then $\angle P S T=180^{\circ}-60^{\circ}=120^{\circ}$.

Therefore, in $\triangle P S T$, we know that $\angle P S T=120^{\circ}$, that $P S=6$ and that $P T=14$.

By the cosine law,

$$
\begin{aligned}
P T^{2} & =P S^{2}+S T^{2}-2(P S)(S T) \cos (\angle P S T) \\
14^{2} & =6^{2}+S T^{2}-2(6)(S T) \cos \left(120^{\circ}\right) \\
196 & =36+S T^{2}+6 S T \quad\left(\text { since } \cos \left(120^{\circ}\right)=-\frac{1}{2}\right) \\
0 & =S T^{2}+6 S T-160 \\
0 & =(S T-10)(S T+16)
\end{aligned}
$$

and so $S T=10$ or $S T=-16$. Since $S T>0$, then $S T=10$.

ANswer: (D)