Question

look at the two triangles below. triangle a: vertices at (1, 1), (3, 1), (2, 3) triangle b: vertices at (4, 1), (6, 1), (5, 3) b) what do you notice about the two triangles? b) do you think they are congruent? why or why not?

Explanation:

Step1: Recall the distance - formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate side - lengths of Triangle A

Let the vertices of Triangle A be $A_1(1,1)$, $A_2(3,1)$, $A_3(2,3)$.
The length of side $A_1A_2$: $d_{A_1A_2}=\sqrt{(3 - 1)^2+(1 - 1)^2}=\sqrt{2^2+0^2}=2$.
The length of side $A_1A_3$: $d_{A_1A_3}=\sqrt{(2 - 1)^2+(3 - 1)^2}=\sqrt{1 + 4}=\sqrt{5}$.
The length of side $A_2A_3$: $d_{A_2A_3}=\sqrt{(2 - 3)^2+(3 - 1)^2}=\sqrt{1 + 4}=\sqrt{5}$.

Step3: Calculate side - lengths of Triangle B

Let the vertices of Triangle B be $B_1(4,1)$, $B_2(6,1)$, $B_3(5,3)$.
The length of side $B_1B_2$: $d_{B_1B_2}=\sqrt{(6 - 4)^2+(1 - 1)^2}=\sqrt{2^2+0^2}=2$.
The length of side $B_1B_3$: $d_{B_1B_3}=\sqrt{(5 - 4)^2+(3 - 1)^2}=\sqrt{1 + 4}=\sqrt{5}$.
The length of side $B_2B_3$: $d_{B_2B_3}=\sqrt{(5 - 6)^2+(3 - 1)^2}=\sqrt{1 + 4}=\sqrt{5}$.

Step4: Compare the side - lengths

The side - lengths of Triangle A are equal to the corresponding side - lengths of Triangle B. That is, $d_{A_1A_2}=d_{B_1B_2}$, $d_{A_1A_3}=d_{B_1B_3}$, and $d_{A_2A_3}=d_{B_2B_3}$.

Step5: Determine congruence

By the Side - Side - Side (SSS) congruence criterion, if the three sides of one triangle are equal in length to the three sides of another triangle, the two triangles are congruent.

Answer:

b) The two triangles have the same side - lengths.
b) Yes, they are congruent. Because the lengths of the corresponding sides of the two triangles are equal ($AB = DE$, $AC=DF$, $BC = EF$), so by the SSS (Side - Side - Side) congruence criterion, the two triangles are congruent.