Question

the graph shows triangles efg and stu.
is efg similar to stu? justify your answer.
yes, because a rotation 90° counterclockwise around the origin maps efg onto stu.
yes, because a translation left 7 units and up 4 units maps efg onto stu.
no, because ∠f and ∠t do not have the same measure.

Explanation:

Step1: List coordinates of vertices

Triangles:

• $EFG$: $E(6, -4)$, $F(3, 1)$, $G(5, 4)$
• $STU$: $S(5, 6)$, $T(1, 3)$, $U(-3, 5)$

Step2: Calculate side lengths (distance formula)

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

• $EFG$ sides:

$FG=\sqrt{(5-3)^2+(4-1)^2}=\sqrt{4+9}=\sqrt{13}$
$FE=\sqrt{(6-3)^2+(-4-1)^2}=\sqrt{9+25}=\sqrt{34}$
$EG=\sqrt{(6-5)^2+(-4-4)^2}=\sqrt{1+64}=\sqrt{65}$

• $STU$ sides:

$TU=\sqrt{(-3-1)^2+(5-3)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$
$TS=\sqrt{(5-1)^2+(6-3)^2}=\sqrt{16+9}=5$
$SU=\sqrt{(5-(-3))^2+(6-5)^2}=\sqrt{64+1}=\sqrt{65}$

Step3: Check angle measures (slopes for right angles)

Slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$

• $EFG$ slopes:

$m_{FG}=\frac{4-1}{5-3}=\frac{3}{2}$, $m_{FE}=\frac{-4-1}{6-3}=-\frac{5}{3}$
Product: $\frac{3}{2} \times -\frac{5}{3}=-\frac{5}{2}
eq -1$; $\angle F$ is not right.

• $STU$ slopes:

$m_{TU}=\frac{5-3}{-3-1}=-\frac{1}{2}$, $m_{TS}=\frac{6-3}{5-1}=\frac{3}{4}$
Product: $-\frac{1}{2} \times \frac{3}{4}=-\frac{3}{8}
eq -1$; but compare $\angle F$ and $\angle T$:
$\angle F$: sides with slopes $\frac{3}{2}, -\frac{5}{3}$; $\angle T$: sides with slopes $-\frac{1}{2}, \frac{3}{4}$. The angle measures differ as the slope pairs do not produce congruent angles.

Step4: Check similarity conditions

Similarity requires all angles congruent AND sides proportional. Side ratios are not equal ($\frac{\sqrt{13}}{2\sqrt{5}}
eq \frac{\sqrt{34}}{5}
eq \frac{\sqrt{65}}{\sqrt{65}}$), and $\angle F
eq \angle T$.

Answer:

No, because $\angle F$ and $\angle T$ do not have the same measure.