Question

Which triangles are congruent?plz help!!!!.

Answer 1

The right answer is triangles ABC and MNO. Congruent triangles are triangles that have the same lengths of its three sides. To get the length of each side, use the distance formula with equation:
$d = \sqrt{( x_{2} - x_{1}) ^{2} +( y_{2} - y_{1} )^2 }$

ABC and MNO have the same segment lengths of its correspoding sides:
AB = MN = 8 units
BC = NO = 6 units
AC = MO = 10 units

Answer 2

Answer:

Option C. is correct

Step-by-step explanation:

In the triangle ABC, vertices are A(12, 8), B(4, 8) and C(4, 14)

AB = $\sqrt{(12-4)^{2}+(8-8)x^{2}}=8$

BC = $\sqrt{(4-4)^{2}+(8-14)^{2}}=6$

AC = $\sqrt{(12-4)^{2}+(8-14)^{2})}=10$

In triangle XYZ vertices are X(6,6), Y(4,12) and Z(10, 14)

XY = $\sqrt{(6-4)^{2}+(6-12)^{2}}=2\sqrt{10}$

YZ = $\sqrt{(4-10)^{2}+(12-14)^{2}}=2\sqrt{10}$

XZ = $\sqrt{(10-6)^{2}+(14-6)^{2}}=4\sqrt{5}$

In triangle MNO, vertices are M(4, 16), N(4, 8) and O(-2, 8)

MN = $\sqrt{(4-4)^{2}+(16-8)^{2}}=8$

MO = $\sqrt{(4+2)^{2}+(16-8)^{2}}=10$

NO = $\sqrt{(4+2)^{2}+(8-8)^{2}}=6$

In triangle JKL, vertices are J(14, -2), K(12,2) and L(20, 4)

JK = $\sqrt{(14-12)^{2}+(-2-2)^{2}}=2\sqrt{5}$

KL = $\sqrt{(20-12)^{2}+(4-2)^{2}}=2\sqrt{17}$

JL = $\sqrt{(20-14)^{2}+(4+2)^{2}}=6\sqrt{2}$

Now we compare the sides for the congruence

Since AB = MN = 8

BC = NO = 6

AC = OM = 10

So ΔABC ≅ Δ MNO

Option C. is correct