Question

the graph shows triangles cde and vwx. is cde congruent to vwx?

Explanation:

Step1: Identify coordinates of vertices

Triangles:

• $CDE$: $C(8,1)$, $D(3,3)$, $E(5,-4)$
• $VWX$: $V(-8,-2)$, $W(-3,-3)$, $X(-5,3)$

Step2: Calculate side lengths (distance formula)

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

For $\triangle CDE$:

$CD=\sqrt{(8-3)^2+(1-3)^2}=\sqrt{25+4}=\sqrt{29}$
$DE=\sqrt{(5-3)^2+(-4-3)^2}=\sqrt{4+49}=\sqrt{53}$
$CE=\sqrt{(8-5)^2+(1-(-4))^2}=\sqrt{9+25}=\sqrt{34}$

For $\triangle VWX$:

$VW=\sqrt{(-3-(-8))^2+(-3-(-2))^2}=\sqrt{25+1}=\sqrt{26}$
$WX=\sqrt{(-5-(-3))^2+(3-(-3))^2}=\sqrt{4+36}=\sqrt{40}$
$VX=\sqrt{(-5-(-8))^2+(3-(-2))^2}=\sqrt{9+25}=\sqrt{34}$

Step3: Compare corresponding side lengths

Only $CE = VX = \sqrt{34}$; $CD
eq VW$, $DE
eq WX$. No full match of all 3 sides.

Answer:

No, $\triangle CDE$ is not congruent to $\triangle VWX$.