Question

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are \\( \triangle xyz \\) and \\( \triangle def \\) congruent?

Explanation:

Step1: Calculate the length of the base

For $\triangle DEF$, the base $ED$: $E(0,2)$ and $D(10,2)$. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, since $y_1 = y_2 = 2$, $d=\vert10 - 0\vert=10$.
For $\triangle XYZ$, the base $YX$: $Y(0,-10)$ and $X(10,-10)$. Using the distance formula, since $y_1 = y_2=-10$, $d=\vert10 - 0\vert=10$.

Step2: Calculate the height

For $\triangle DEF$, the height: $F(5,10)$ and the base $ED$ (where $y = 2$). The height $h=\vert10 - 2\vert=8$.
For $\triangle XYZ$, the height: $Z(5,-2)$ and the base $YX$ (where $y=-10$). The height $h=\vert-2-(-10)\vert = 8$.

Step3: Check congruence

Since both triangles have base $b = 10$ and height $h = 8$, using the formula for the area of a triangle $A=\frac{1}{2}bh$ (they have the same area). Also, by the SSS (if we calculate the side - lengths using the distance formula:
For $\triangle DEF$:

• $ED=\sqrt{(10 - 0)^2+(2 - 2)^2}=10$
• $EF=\sqrt{(5 - 0)^2+(10 - 2)^2}=\sqrt{25 + 64}=\sqrt{89}$
• $FD=\sqrt{(10 - 5)^2+(2 - 10)^2}=\sqrt{25 + 64}=\sqrt{89}$

For $\triangle XYZ$:

• $YX=\sqrt{(10 - 0)^2+(-10+10)^2}=10$
• $YZ=\sqrt{(5 - 0)^2+(-2 + 10)^2}=\sqrt{25 + 64}=\sqrt{89}$
• $ZX=\sqrt{(10 - 5)^2+(-10 + 2)^2}=\sqrt{25+64}=\sqrt{89}$

Since $ED = YX$, $EF = YZ$, $FD=ZX$, by SSS (Side - Side - Side) congruence criterion, $\triangle XYZ\cong\triangle DEF$.

Answer:

Yes, $\triangle XYZ$ and $\triangle DEF$ are congruent.