Question

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are △def and △abc congruent?

Explanation:

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

Using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(AB\): \(A(-8,-10)\), \(B(-8,-1)\)
\(AB=\sqrt{(-8 + 8)^2+(-1+ 10)^2}=\sqrt{0 + 81}=9\)
For \(BC\): \(B(-8,-1)\), \(C(2,-1)\)
\(BC=\sqrt{(2 + 8)^2+(-1+ 1)^2}=\sqrt{100+0}=10\)
For \(AC\): \(A(-8,-10)\), \(C(2,-1)\)
\(AC=\sqrt{(2 + 8)^2+(-1 + 10)^2}=\sqrt{100 + 81}=\sqrt{181}\)

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

For \(DE\): \(D(0,8)\), \(E(10,8)\)
\(DE=\sqrt{(10 - 0)^2+(8 - 8)^2}=\sqrt{100+0}=10\)
For \(EF\): \(E(10,8)\), \(F(10,-1)\)
\(EF=\sqrt{(10 - 10)^2+(-1 - 8)^2}=\sqrt{0+81}=9\)
For \(DF\): \(D(0,8)\), \(F(10,-1)\)
\(DF=\sqrt{(10 - 0)^2+(-1 - 8)^2}=\sqrt{100 + 81}=\sqrt{181}\)

Step3: Check the congruence

Since \(AB = EF = 9\), \(BC=DE = 10\), \(AC = DF=\sqrt{181}\)
By SSS (Side - Side - Side) congruence criterion, \(\triangle ABC\cong\triangle EFD\)

Answer:

Yes, \(\triangle DEF\) and \(\triangle ABC\) are congruent.