SAT Question 5b4757df: Lines, Angles, and Triangles answer and explanation with Desmos solution

It's given that in triangle $RST$, angle $T$ is a right angle. The area of a right triangle can be found using the formula $A=\frac{1}{2}{l}_1{l}_2$, where $A$ represents the area of the right triangle, $l_1$ represents the length of one leg of the triangle, and $l_2$ represents the length of the other leg of the triangle. In triangle $RST$, the two legs are $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ and $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$. Therefore, if the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $72$ and the area of triangle $RST$ is $792$, then $792=\frac{1}{2}\left(72\right)\left(ST\right)$, or $792=\left(36\right)\left(ST\right)$. Dividing both sides of this equation by $36$ yields $22=ST$. Therefore, the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $22$. It's also given that point $L$ lies on $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$, point $K$ lies on $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$, and $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is parallel to $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$. It follows that angle $LKS$ is a right angle. Since triangles $RST$ and $LSK$ share angle $S$ and have right angles $T$ and $K$, respectively, triangles $RST$ and $LSK$ are similar triangles. Therefore, the ratio of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ to the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is equal to the ratio of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ to the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$. If the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $72$ and the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $24$, it follows that the ratio of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ to the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $\frac{72}{24}$, or $3$, so the ratio of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ to the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $3$. Therefore, $\frac{22}{SK}=3$. Multiplying both sides of this equation by $SK$ yields $22=\left(3\right)\left(SK\right)$. Dividing both sides of this equation by $3$ yields $\frac{22}{3}=SK$. Since the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$, $22$, is the sum of the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$, $\frac{22}{3}$, and the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$, it follows that the length of $PARS{E}_{E}RROR:Entit{y}^{\prime }mac{r}^{\prime }notdefined,line1,column139(<string>,line1)$ is $22-\frac{22}{3}$, or $\frac{44}{3}$. Note that 44/3, 14.66, and 14.67 are examples of ways to enter a correct answer.