Question

congruent triangles (topic 5)
which triangle is congruent to △klm?
none
△gjh
△cba
△fde

Explanation:

Step1: Recall congruent - triangle criteria

Two triangles are congruent if they have the same size and shape, which means corresponding sides and corresponding angles are equal. We can use the distance - formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ to find the lengths of the sides of the triangles.
Let the coordinates of the vertices of $\triangle KLM$ be $K(-2,2)$, $L(-4,1)$, $M(-2, - 1)$.
The length of $KL=\sqrt{(-2+4)^2+(2 - 1)^2}=\sqrt{4 + 1}=\sqrt{5}$.
The length of $KM=\sqrt{(-2 + 2)^2+(2+1)^2}=3$.
The length of $LM=\sqrt{(-4 + 2)^2+(1 + 1)^2}=\sqrt{4+4}=2\sqrt{2}$.

Step2: Find side - lengths of $\triangle GJH$

Let $G(4,2)$, $J(2,0)$, $H(2,2)$.
$GJ=\sqrt{(4 - 2)^2+(2 - 0)^2}=\sqrt{4 + 4}=2\sqrt{2}$.
$GH=\sqrt{(4 - 2)^2+(2 - 2)^2}=2$.
$JH=\sqrt{(2 - 2)^2+(2 - 0)^2}=2$.

Step3: Find side - lengths of $\triangle CBA$

Let $C(2,-2)$, $B(0,-4)$, $A(2,-4)$.
$CB=\sqrt{(2 - 0)^2+(-2 + 4)^2}=\sqrt{4 + 4}=2\sqrt{2}$.
$CA=\sqrt{(2 - 2)^2+(-2 + 4)^2}=2$.
$BA=\sqrt{(0 - 2)^2+(-4 + 4)^2}=2$.

Step4: Find side - lengths of $\triangle FDE$

Let $F(4,0)$, $D(4,-2)$, $E(6,-2)$.
$FD=\sqrt{(4 - 4)^2+(0 + 2)^2}=2$.
$FE=\sqrt{(4 - 6)^2+(0 + 2)^2}=\sqrt{4 + 4}=2\sqrt{2}$.
$DE=\sqrt{(4 - 6)^2+(-2+2)^2}=2$.

Step5: Compare side - lengths

We can also use the concept of rigid motions (translations, rotations, and reflections). By observing the grid, we can see that $\triangle KLM$ can be translated and reflected to match $\triangle CBA$.
We can translate $\triangle KLM$ 4 units to the right and 4 units down. Then, we may need to perform a reflection. The side - lengths of $\triangle KLM$ and $\triangle CBA$ match:
$KL$ corresponds to $CB$, $KM$ corresponds to $CA$, and $LM$ corresponds to $BA$.

Answer:

C. $\triangle CBA$