Question

1. which triangles are congruent? enter your answer. check answer

Explanation:

Step1: Use distance - formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Calculate the side - lengths of each triangle.
For $\triangle ABC$:
Let $A(-4,4)$, $B(-2,4)$, $C(-2,2)$.
$AB=\sqrt{(-2 + 4)^2+(4 - 4)^2}=\sqrt{4+0}=2$;
$BC=\sqrt{(-2 + 2)^2+(2 - 4)^2}=\sqrt{0 + 4}=2$;
$AC=\sqrt{(-2 + 4)^2+(2 - 4)^2}=\sqrt{4 + 4}=2\sqrt{2}$.
For $\triangle JHG$:
Let $J(2,4)$, $H(2,2)$, $G(4,2)$.
$JH=\sqrt{(2 - 2)^2+(4 - 2)^2}=\sqrt{0+4}=2$;
$HG=\sqrt{(4 - 2)^2+(2 - 2)^2}=\sqrt{4+0}=2$;
$JG=\sqrt{(4 - 2)^2+(4 - 2)^2}=\sqrt{4 + 4}=2\sqrt{2}$.
For $\triangle DEF$:
Let $D(-4,-4)$, $E(-2,-4)$, $F(-4,-2)$.
$DE=\sqrt{(-2 + 4)^2+(-4 + 4)^2}=\sqrt{4+0}=2$;
$EF=\sqrt{(-4 + 2)^2+(-2 + 4)^2}=\sqrt{4 + 4}=2\sqrt{2}$;
$DF=\sqrt{(-4 + 4)^2+(-2 + 4)^2}=\sqrt{0+4}=2$.
For $\triangle KLM$:
Let $K(4,-4)$, $L(2,-4)$, $M(2,-2)$.
$KL=\sqrt{(2 - 4)^2+(-4 + 4)^2}=\sqrt{4+0}=2$;
$LM=\sqrt{(2 - 2)^2+(-2 + 4)^2}=\sqrt{0 + 4}=2$;
$KM=\sqrt{(2 - 4)^2+(-2 + 4)^2}=\sqrt{4 + 4}=2\sqrt{2}$.

Step2: Compare side - lengths

By comparing the side - lengths of the four triangles, we find that $\triangle ABC\cong\triangle JHG\cong\triangle DEF\cong\triangle KLM$ (by SSS congruence criterion, since all three pairs of corresponding sides are equal).

Answer:

$\triangle ABC\cong\triangle JHG\cong\triangle DEF\cong\triangle KLM$