Question

which statement proves fgh is similar to rst?
$\frac{gh}{fh}=\frac{st}{rs}$
$\frac{fg}{fh}=\frac{st}{rt}$
$\frac{fh}{fg}=\frac{st}{rt}$
$\frac{fg}{gh}=\frac{rs}{st}$

Explanation:

Step1: Identify coordinates of points

Points for $\triangle FGH$: $G(-4,0)$, $H(0,0)$, $F(-4,-4)$
Points for $\triangle RST$: $R(3,2)$, $S(3,4)$, $T(5,4)$

Step2: Calculate side lengths (FGH)

• $FG$: Vertical distance, $|0 - (-4)| = 4$
• $GH$: Horizontal distance, $|0 - (-4)| = 4$
• $FH$: Distance formula: $\sqrt{(-4+4)^2 + (-4-0)^2} = 4\sqrt{2}$

Step3: Calculate side lengths (RST)

• $RS$: Vertical distance, $|4 - 2| = 2$
• $ST$: Horizontal distance, $|5 - 3| = 2$
• $RT$: Distance formula: $\sqrt{(5-3)^2 + (4-2)^2} = 2\sqrt{2}$

Step4: Test each ratio option

• Option1: $\frac{GH}{FH}=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}$; $\frac{ST}{RS}=\frac{2}{2}=1$ → Not equal
• Option2: $\frac{FG}{FH}=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}$; $\frac{ST}{RT}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$ → Equal
• Option3: $\frac{FH}{FG}=\frac{4\sqrt{2}}{4}=\sqrt{2}$; $\frac{ST}{RT}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$ → Not equal
• Option4: $\frac{FG}{GH}=\frac{4}{4}=1$; $\frac{RS}{ST}=\frac{2}{2}=1$ → Equal, but mismatched corresponding sides (right angles are at $G$ and $S$, so $FG$ corresponds to $RS$, $GH$ corresponds to $ST$)

Answer:

$\boldsymbol{\frac{FG}{FH} = \frac{ST}{RT}}$ (the second option)