Question

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are \\( \triangle vwx \\) and \\( \triangle pqr \\) congruent?

Explanation:

Step1: Calculate the side lengths of \(\triangle VWX\)

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(VX\): \(V(0,9)\), \(X(9,0)\)
\(VX=\sqrt{(9 - 0)^2+(0 - 9)^2}=\sqrt{81 + 81}=\sqrt{162}=9\sqrt{2}\)
For \(VW\): \(V(0,9)\), \(W(9,9)\)
\(VW=\sqrt{(9 - 0)^2+(9 - 9)^2}=\sqrt{81+0}=9\)
For \(WX\): \(W(9,9)\), \(X(9,0)\)
\(WX=\sqrt{(9 - 9)^2+(0 - 9)^2}=\sqrt{0 + 81}=9\)

Step2: Calculate the side lengths of \(\triangle PQR\)

For \(PQ\): \(P(1,-9)\), \(Q(-9,-9)\)
\(PQ=\sqrt{(-9 - 1)^2+(-9+9)^2}=\sqrt{100 + 0}=10\)
For \(QR\): \(Q(-9,-9)\), \(R(-9,1)\)
\(QR=\sqrt{(-9 + 9)^2+(1 + 9)^2}=\sqrt{0+100}=10\)
For \(PR\): \(P(1,-9)\), \(R(-9,1)\)
\(PR=\sqrt{(-9 - 1)^2+(1 + 9)^2}=\sqrt{100+100}=\sqrt{200}=10\sqrt{2}\)

Step3: Compare the side lengths

Since \(VX = 9\sqrt{2}
eq10\sqrt{2}=PR\), \(VW = 9
eq10 = PQ\), \(WX = 9
eq10=QR\)

Answer:

No, \(\triangle VWX\) and \(\triangle PQR\) are not congruent.