Question

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δabc has vertices at a(11, 6), b(5, 6), and c(5, 17).
δxyz has vertices at x(-10, 5), y(-12, -2), and z(-4, 15).
δmno has vertices at m(-9, -4), n(-3, -4), and o(-3, -15).
δjkl has vertices at j(17, -2), k(12, -2), and l(12, 7).
δpqr has vertices at p(12, 3), q(12, -2), and r(3, -2).

can be shown to be congruent by a sequence of reflections and translations.
can be shown to be congruent by a single rotation.

Explanation:

Step1: Recall congruence criteria

Two triangles are congruent if they have the same shape and size. Transformations like reflections, translations, and rotations preserve congruence.

Step2: Calculate side - lengths

For \(\triangle ABC\) with \(A(11,6)\), \(B(5,6)\), and \(C(5,17)\):
The length of \(AB=\vert11 - 5\vert=6\) (since \(y\) - coordinates of \(A\) and \(B\) are the same), \(BC=\vert17 - 6\vert = 11\) (since \(x\) - coordinates of \(B\) and \(C\) are the same), and \(AC=\sqrt{(11 - 5)^2+(17 - 6)^2}=\sqrt{36 + 121}=\sqrt{157}\).
For \(\triangle XYZ\) with \(X(-10,5)\), \(Y(-12,-2)\), and \(Z(-4,15)\):
\(XY=\sqrt{(-10+12)^2+(5 + 2)^2}=\sqrt{4 + 49}=\sqrt{53}\), \(YZ=\sqrt{(-12 + 4)^2+(-2 - 15)^2}=\sqrt{64+289}=\sqrt{353}\), \(XZ=\sqrt{(-10 + 4)^2+(5 - 15)^2}=\sqrt{36 + 100}=\sqrt{136}\).
For \(\triangle MNO\) with \(M(-9,-4)\), \(N(-3,-4)\), and \(O(-3,-15)\):
\(MN=\vert-9+3\vert = 6\), \(NO=\vert-15 + 4\vert=11\), \(MO=\sqrt{(-9 + 3)^2+(-4 + 15)^2}=\sqrt{36+121}=\sqrt{157}\).
For \(\triangle JKL\) with \(J(17,-2)\), \(K(12,-2)\), and \(L(12,7)\):
\(JK=\vert17 - 12\vert=5\), \(KL=\vert7 + 2\vert = 9\), \(JL=\sqrt{(17 - 12)^2+(7 + 2)^2}=\sqrt{25 + 81}=\sqrt{106}\).
For \(\triangle PQR\) with \(P(12,3)\), \(Q(12,-2)\), and \(R(3,-2)\):
\(PQ=\vert3 + 2\vert=5\), \(QR=\vert12 - 3\vert = 9\), \(PR=\sqrt{(12 - 3)^2+(3 + 2)^2}=\sqrt{81+25}=\sqrt{106}\).

Step3: Determine congruent triangles by transformation

• \(\triangle ABC\) and \(\triangle MNO\):

The side - lengths of \(\triangle ABC\) and \(\triangle MNO\) are equal (\(AB = MN = 6\), \(BC=NO = 11\), \(AC = MO=\sqrt{157}\)). A sequence of reflections and translations can map \(\triangle ABC\) onto \(\triangle MNO\).

• \(\triangle JKL\) and \(\triangle PQR\):

The side - lengths of \(\triangle JKL\) and \(\triangle PQR\) are equal (\(JK = PQ = 5\), \(KL=QR = 9\), \(JL = PR=\sqrt{106}\)). A single rotation can map \(\triangle JKL\) onto \(\triangle PQR\).

Answer:

1. \(\triangle ABC\) and \(\triangle MNO\)
2. \(\triangle JKL\) and \(\triangle PQR\)