Question

plot the triangles, △abc and △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 △abc ~ △def show your work here hint: to add the square root symbol (√□), type
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Explanation:

Step1: Calculate length of AB

Use distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. For A(-5, -2) and B(-3, 4):
$AB = \sqrt{(-3 - (-5))^2 + (4 - (-2))^2} = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$

Step2: Calculate length of BC

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

Step3: Calculate length of AC

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

Step4: Calculate length of DE

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

Step5: Calculate length of EF

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

Step6: Calculate length of DF

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

Step7: Find ratios of corresponding sides

$\frac{AB}{DE} = \frac{2\sqrt{10}}{5\sqrt{10}} = \frac{2}{5}$, $\frac{BC}{EF} = \frac{4\sqrt{2}}{10\sqrt{2}} = \frac{2}{5}$, $\frac{AC}{DF} = \frac{2\sqrt{10}}{5\sqrt{10}} = \frac{2}{5}$
Since all three ratios are equal, by SSS similarity criterion, $\triangle ABC \sim \triangle DEF$.

Answer:

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