Question

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

Explanation:

Step1: Find the lengths of the sides of triangle CDE

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(CD\): \(C(8,2)\), \(D(4,3)\)
\(CD=\sqrt{(8 - 4)^2+(2 - 3)^2}=\sqrt{16 + 1}=\sqrt{17}\)
For \(DE\): \(D(4,3)\), \(E(5,-3)\)
\(DE=\sqrt{(5 - 4)^2+(-3 - 3)^2}=\sqrt{1+36}=\sqrt{37}\)
For \(CE\): \(C(8,2)\), \(E(5,-3)\)
\(CE=\sqrt{(8 - 5)^2+(2+3)^2}=\sqrt{9 + 25}=\sqrt{34}\)

Step2: Find the lengths of the sides of triangle VWX

For \(VW\): \(V(-8,-2)\), \(W(-4,-3)\)
\(VW=\sqrt{(-4 + 8)^2+(-3 + 2)^2}=\sqrt{16+1}=\sqrt{17}\)
For \(WX\): \(W(-4,-3)\), \(X(-6,3)\)
\(WX=\sqrt{(-6 + 4)^2+(3 + 3)^2}=\sqrt{4 + 36}=\sqrt{40}\)
For \(VX\): \(V(-8,-2)\), \(X(-6,3)\)
\(VX=\sqrt{(-6 + 8)^2+(3 + 2)^2}=\sqrt{4+25}=\sqrt{29}\)

Step3: Compare the side lengths

Since the side lengths of \(\triangle CDE\) (\(\sqrt{17},\sqrt{37},\sqrt{34}\)) and \(\triangle VWX\) (\(\sqrt{17},\sqrt{40},\sqrt{29}\)) are not all equal.

Answer:

No, \(\triangle CDE\) is not congruent to \(\triangle VWX\) because their corresponding side lengths are not all equal.