Question

plot the triangles, $\triangle abc$ and $\triangle def$ with vertices\
$a(-5, -2)\\ \\ \\ b(-3, 4)\\ \\ \\ c(1, 0)$\
$d(-8, -5)\\ \\ \\ e(-3, 10)\\ \\ \\ f(7, 0)$\
then, find the measurements below and use them to prove if\
$\triangle abc \sim \triangle def$\
\
show your work here\
\
hint to add the square root symbol ($\sqrt{\square}$), type
oot\

Explanation:

Step1: Calculate lengths of sides of $\triangle ABC$

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

• Length of $AB$:

$A(-5,-2), B(-3,4)$
$AB = \sqrt{(-3 - (-5))^2 + (4 - (-2))^2} = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$

• Length of $BC$:

$B(-3,4), C(1,0)$
$BC = \sqrt{(1 - (-3))^2 + (0 - 4)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

• Length of $AC$:

$A(-5,-2), C(1,0)$
$AC = \sqrt{(1 - (-5))^2 + (0 - (-2))^2} = \sqrt{(6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$

Step2: Calculate lengths of sides of $\triangle DEF$

• Length of $DE$:

$D(-8,-5), E(-3,10)$
$DE = \sqrt{(-3 - (-8))^2 + (10 - (-5))^2} = \sqrt{(5)^2 + (15)^2} = \sqrt{25 + 225} = \sqrt{250} = 5\sqrt{10}$

• Length of $EF$:

$E(-3,10), F(7,0)$
$EF = \sqrt{(7 - (-3))^2 + (0 - 10)^2} = \sqrt{(10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}$

• Length of $DF$:

$D(-8,-5), F(7,0)$
$DF = \sqrt{(7 - (-8))^2 + (0 - (-5))^2} = \sqrt{(15)^2 + (5)^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10}$

Step3: Check the ratios of corresponding sides

• Ratio of $AB$ to $DE$: $\frac{AB}{DE} = \frac{2\sqrt{10}}{5\sqrt{10}} = \frac{2}{5}$
• Ratio of $BC$ to $EF$: $\frac{BC}{EF} = \frac{4\sqrt{2}}{10\sqrt{2}} = \frac{4}{10} = \frac{2}{5}$
• Ratio of $AC$ to $DF$: $\frac{AC}{DF} = \frac{2\sqrt{10}}{5\sqrt{10}} = \frac{2}{5}$

Since the ratios of corresponding sides are equal, by the SSS (Side - Side - Side) similarity criterion, $\triangle ABC \sim \triangle DEF$

Answer:

$\triangle ABC \sim \triangle DEF$ (by SSS similarity criterion as the ratios of corresponding sides $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{2}{5}$)