Question

are $\triangle vwx$ and $\triangle fgh$ congruent?
yes no

Explanation:

Step1: Find coordinates of vertices

For $\triangle VWX$: $V(0,0)$, $W(-6,-4)$, $X(-2,-10)$
For $\triangle FGH$: $F(8,0)$, $G(10,7)$, $H(1,8)$

Step2: Calculate side lengths (distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$)

- $VW$: $\sqrt{(-6-0)^2+(-4-0)^2}=\sqrt{36+16}=\sqrt{52}$
- $WX$: $\sqrt{(-2+6)^2+(-10+4)^2}=\sqrt{16+36}=\sqrt{52}$
- $VX$: $\sqrt{(-2-0)^2+(-10-0)^2}=\sqrt{4+100}=\sqrt{104}$

- $FG$: $\sqrt{(10-8)^2+(7-0)^2}=\sqrt{4+49}=\sqrt{53}$ (Wait, correction: $G(10,7)$, $F(8,0)$: $\sqrt{(10-8)^2+(7-0)^2}=\sqrt{4 + 49}=\sqrt{53}$? No, earlier $VW$ was $\sqrt{52}$, so miscalculation. Wait, recheck $H(1,8)$, $F(8,0)$: $HF$: $\sqrt{(8-1)^2+(0-8)^2}=\sqrt{49+64}=\sqrt{113}$? No, better to re-express coordinates correctly.

Wait, correct coordinates:
$V(0,0)$, $W(-5,-4)$? Wait grid: each square is 1 unit. Let's re-identify:
$V$ is at (0,0). $W$: left 5, down 4? Wait the graph: $W$ is at (-5, -4)? Wait original image: $W$ is at (-6, -4)? No, looking at the grid, $W$ is at (-5, -4)? Wait maybe I misread. Let's do properly:

For $\triangle VWX$:
- $V$: (0, 0)
- $W$: (-5, -4) (since from V, left 5, down 4)
- $X$: (-2, -10) (left 2, down 10)

For $\triangle FGH$:
- $F$: (8, 0)
- $G$: (10, 7) (right 2 from F, up 7)
- $H$: (1, 8) (left 7 from F, up 8)

Now calculate sides:

$VW$: distance between (0,0) and (-5,-4): $\sqrt{(-5-0)^2 + (-4-0)^2} = \sqrt{25 + 16} = \sqrt{41}$? No, this is confusing. Alternative: check if triangles have same side lengths (SSS) or congruent via translation/rotation.

Wait, maybe the triangles are congruent by SSS. Wait let's use correct grid:

$V(0,0)$, $W(-5, -4)$, $X(-2, -10)$? No, $X$ is at (-2, -10)? Wait the vertical axis: $X$ is at y=-10? No, the y-axis goes down to -10? Wait the graph has y from -10 to 10.

Wait, another approach: count the lengths of sides by counting grid units (horizontal/vertical differences, then use Pythagoras).

For $\triangle VWX$:
- $VW$: horizontal difference: $| -5 - 0 | = 5$, vertical difference: $| -4 - 0 | = 4$, so length $\sqrt{5^2 + 4^2} = \sqrt{41}$? No, maybe $W$ is at (-6, -4), $X$ at (-2, -10). Then $VW$: from (0,0) to (-6,-4): $\sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$
$WX$: from (-6,-4) to (-2,-10): horizontal +4, vertical -6: $\sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52}$
$VX$: from (0,0) to (-2,-10): $\sqrt{(-2)^2 + (-10)^2} = \sqrt{4 + 100} = \sqrt{104}$

For $\triangle FGH$:
$F(8,0)$, $G(10,7)$, $H(1,8)$? No, $H$ is at (1,8), $F(8,0)$, $G(10,7)$

$HF$: from (1,8) to (8,0): horizontal +7, vertical -8: $\sqrt{7^2 + (-8)^2} = \sqrt{49 + 64} = \sqrt{113}$ – no, not matching. Wait, maybe $H$ is at (1,8), $G(10,7)$, $F(8,0)$:

$FG$: (10,7) to (8,0): horizontal -2, vertical -7: $\sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53}$ – not $\sqrt{52}$.

Wait, I must have misidentified coordinates. Let's look again: the blue triangle $\triangle FGH$: $H$ is at (1,8), $G$ at (10,7), $F$ at (8,0). The brown triangle $\triangle VWX$: $V$ at (0,0), $W$ at (-5,-4), $X$ at (-2,-10).

Wait, maybe the correct way is to check if the triangles have the same side lengths. Let's use vectors or translation. Wait, maybe the triangles are congruent because their corresponding sides are equal. Wait, let's recalculate $VW$: distance from V(0,0) to W(-5,-4): $\sqrt{25 + 16} = \sqrt{41}$. No, this is wrong.

Alternative: the key is that congruent triangles have all corresponding sides equal. Let's use the distance formula correctly.

Correct coordinates (assuming each grid square is 1 unit):

- $\triangle VWX$:
$V$: (0, 0)
$W$: (-5, -4) (5 left, 4 down)
$X$: (-2, -10) (2 left, 10 down)

- $\triangle FGH$:
$F$: (8, 0)
$G$: (10, 7) (2 right, 7 up)
$H$: (1, 8) (7 left, 8 up)

Now calculate side lengths:

$VW$: $\sqrt{(-5 - 0)^2 + (-4 - 0)^2} = \sqrt{25 + 16} = \sqrt{41}$
$WX$: $\sqrt{(-2 - (-5))^2 + (-10 - (-4))^2} = \sqrt{(3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45}$
$VX$: $\sqrt{(-2 - 0)^2 + (-10 - 0)^2} = \sqrt{4 + 100} = \sqrt{104}$

$FG$: $\sqrt{(10 - 8)^2 + (7 - 0)^2} = \sqrt{4 + 49} = \sqrt{53}$
$GH$: $\sqrt{(1 - 10)^2 + (8 - 7)^2} = \sqrt{81 + 1} = \sqrt{82}$
$HF$: $\sqrt{(8 - 1)^2 + (0 - 8)^2} = \sqrt{49 + 64} = \sqrt{113}$

These don't match. Wait, I must have misread the coordinates. Let's look at the y-axis: $H$ is at (1,8), $G$ at (10,7), $F$ at (8,0). $V$ at (0,0), $W$ at (-5,-4), $X$ at (-2,-10). No, this can't be. Wait maybe $X$ is at (-2, -9)? No, the grid has y from -10 to 10.

Wait, another approach: the problem is to check congruence, so maybe the triangles are congruent by SSS. Let's re-express $H$ as (1,8), $F$ as (8,0), $G$ as (10,7). $V$ as (0,0), $W$ as (-5,-4), $X$ as (-2,-10). No, the lengths don't match. Wait, maybe I made a mistake in $W$'s coordinates. Let's see: $W$ is at (-5, -4), $V$ at (0,0): distance $\sqrt{25 + 16} = \sqrt{41}$. $F$ at (8,0), $G$ at (10,7): distance $\sqrt{4 + 49} = \sqrt{53}$. Not equal. So the answer is no? Wait, but maybe I misread the graph. Let's check the original question: the blue triangle is $\triangle FGH$, brown is $\triangle VWX$. Let's count the grid units for each side.

Wait, $VW$: from (0,0) to (-5,-4): 5 right? No, left 5, down 4. $FG$: from (8,0) to (10,7): right 2, up 7. Not same. $WX$: from (-5,-4) to (-2,-10): right 3, down 6. $GH$: from (10,7) to (1,8): left 9, up 1. Not same. $VX$: from (0,0) to (-2,-10): left 2, down 10. $HF$: from (1,8) to (8,0): right 7, down 8. Not same. So the triangles are not congruent.

Answer:

no