Question

plot the triangles, $\triangle tuv$ and $\triangle xuy$ with vertices $t(-5, 2)$ $u(-2, 2)$ $v(-2, 7)$ $x(4, 2)$ $y(-2, -8)$ then, find the measurements below and use them to prove if $\triangle tuv \sim \triangle xuy$ show your work here hint to add the square root symbol ($\sqrt{\square}$), type
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Explanation:

Step1: Find lengths of sides in $\triangle TUV$

For $TU$: $T(-5,2)$, $U(-2,2)$. Distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. So $TU = \sqrt{(-2 - (-5))^2 + (2 - 2)^2} = \sqrt{3^2 + 0^2} = 3$.
For $UV$: $U(-2,2)$, $V(-2,7)$. $UV = \sqrt{(-2 - (-2))^2 + (7 - 2)^2} = \sqrt{0^2 + 5^2} = 5$.
For $TV$: $T(-5,2)$, $V(-2,7)$. $TV = \sqrt{(-2 - (-5))^2 + (7 - 2)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$.

Step2: Find lengths of sides in $\triangle XUY$

For $XU$: $X(4,2)$, $U(-2,2)$. $XU = \sqrt{(-2 - 4)^2 + (2 - 2)^2} = \sqrt{(-6)^2 + 0^2} = 6$.
For $UY$: $U(-2,2)$, $Y(-2,-8)$. $UY = \sqrt{(-2 - (-2))^2 + (-8 - 2)^2} = \sqrt{0^2 + (-10)^2} = 10$.
For $XY$: $X(4,2)$, $Y(-2,-8)$. $XY = \sqrt{(-2 - 4)^2 + (-8 - 2)^2} = \sqrt{(-6)^2 + (-10)^2} = \sqrt{36 + 100} = \sqrt{136} = 2\sqrt{34}$.

Step3: Check similarity ratios

$\frac{TU}{XU} = \frac{3}{6} = \frac{1}{2}$, $\frac{UV}{UY} = \frac{5}{10} = \frac{1}{2}$, $\frac{TV}{XY} = \frac{\sqrt{34}}{2\sqrt{34}} = \frac{1}{2}$. All ratios are equal, so $\triangle TUV \sim \triangle XUY$ by SSS similarity.

Answer:

$\triangle TUV \sim \triangle XUY$ (proven by SSS similarity with ratio $\frac{1}{2}$ for corresponding sides).