Square $P Q R S$ has side length 30 , as shown. The square is divided into 5 regions of equal area: $\triangle S P T, \triangle S T U$, $\triangle S V W, \triangle S W R$, and quadrilateral $S U Q V$. The value of $\frac{S U}{S T}$ is closest to
(A) 1.17
(B) 1.19
(C) 1.21
(D) 1.23
(E) 1.25

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-025.jpg?height=368&width=390&top_left_y=648&top_left_x=1301)
The area of square $P Q R S$ is $(30)(30)=900$.

Each of the 5 regions has equal area, and so the area of each region is $900 \div 5=180$.

The area of $\triangle S P T$ is equal to $\frac{1}{2}(P S)(P T)=\frac{1}{2}(30)(P T)=15(P T)$.

The area of $\triangle S P T$ is 180 , and so $15(P T)=180$ or $P T=180 \div 15=12$.

The area of $\triangle S T U$ is 180 .

Let the base of $\triangle S T U$ be $U T$.

The height of $\triangle S T U$ is equal to $P S$ since $P S$ is the perpendicular distance between base $U T$ (extended) and the vertex $S$.

The area of $\triangle S T U$ is equal to $\frac{1}{2}(P S)(U T)=\frac{1}{2}(30)(U T)=15(U T)$.

The area of $\triangle S T U$ is 180 , and so $15(U T)=180$ or $U T=180 \div 15=12$.

In $\triangle S P T, \angle S P T=90^{\circ}$. By the Pythagorean Theorem, $S T^{2}=P S^{2}+P T^{2}$ or $S T^{2}=30^{2}+12^{2}$ or $S T^{2}=900+144=1044$, and so $S T=\sqrt{1044}$ (since $S T>0$ ).

In $\triangle S P U, \angle S P U=90^{\circ}$ and $P U=P T+U T=12+12=24$.

By the Pythagorean Theorem, $S U^{2}=P S^{2}+P U^{2}$ or $S T^{2}=30^{2}+24^{2}$ or $S U^{2}=900+576=1476$, and so $S U=\sqrt{1476}$ (since $S U>0$ ).

Therefore, $\frac{S U}{S T}=\frac{\sqrt{1476}}{\sqrt{1044}}$ which is approximately equal to 1.189.

Of the answers given, $\frac{S U}{S T}$ is closest to 1.19.

ANswer: (B)

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