Question

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

Explanation:

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

Using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(QR\): \(Q(-1,-2)\), \(R(9,-2)\)
\(QR=\sqrt{(9 - (-1))^2+(-2-(-2))^2}=\sqrt{10^2 + 0^2}=10\)
For \(PQ\): \(P(2,5)\), \(Q(-1,-2)\)
\(PQ=\sqrt{(2 - (-1))^2+(5 - (-2))^2}=\sqrt{3^2+7^2}=\sqrt{9 + 49}=\sqrt{58}\)
For \(PR\): \(P(2,5)\), \(R(9,-2)\)
\(PR=\sqrt{(9 - 2)^2+(-2 - 5)^2}=\sqrt{7^2+(-7)^2}=\sqrt{49+49}=\sqrt{98}\)

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

For \(GH\): \(G(-1,4)\), \(H(1,10)\)
\(GH=\sqrt{(1 - (-1))^2+(10 - 4)^2}=\sqrt{2^2+6^2}=\sqrt{4 + 36}=\sqrt{40}\)
For \(GI\): \(G(-1,4)\), \(I(-9,10)\)
\(GI=\sqrt{(-9 - (-1))^2+(10 - 4)^2}=\sqrt{(-8)^2+6^2}=\sqrt{64 + 36}=\sqrt{100}=10\)
For \(HI\): \(H(1,10)\), \(I(-9,10)\)
\(HI=\sqrt{(-9 - 1)^2+(10 - 10)^2}=\sqrt{(-10)^2+0^2}=10\)

Since the side lengths of \(\triangle PQR\) (\(10,\sqrt{58},\sqrt{98}\)) and \(\triangle GHI\) (\(10,\sqrt{40},10\)) are not all equal.

Answer:

\(\triangle PQR\) and \(\triangle GHI\) are not congruent.