CONGRUENT TRIANGLES ON A COORDINATE PLANE WORKSHEET

Congruent Triangles on a Coordinate Plane Worksheet :

Worksheet given in this section will be much useful for the students who would like to practice problems on congruence triangles on the coordinate plane.

Congruent Triangles on a Coordinate Plane Worksheet - Problems

Problem 1 :

In the diagram given below, prove that ΔABC  ≅  ΔFGH

Problem 2 :

In the diagram given below, prove that ΔABC  ≅  ΔDEF

Problem 3 :

In the diagram given below, prove that ΔOPM  ≅  ΔMNP

Congruent Triangles on a Coordinate Plane Worksheet - Solution

Problem 1 :

In the diagram given below, prove that ΔABC  ≅  ΔFGH

Solution :

Because AB = 5 in triangle ABC and FG = 5 in triangle FGH,

AB  ≅  FG.

Because AC = 3 in triangle ABC and FH = 3 in triangle FGH,

AC  ≅  FH.

Use the distance formula to find the lengths of BC and GH.

Length of BC :

BC  =  √[(x₂ - x₁)² + (y₂ - y₁)²]

Here (x₁, y₁)  =  B(-7, 0) and (x₂, y₂)  =  C(-4, 5)

BC  =  √[(-4 + 7)² + (5 - 0)²]

BC  =  √[3² + 5²]

BC  =  √[9 + 25]

BC  =  √34

Length of GH :

GH  =  √[(x₂ - x₁)² + (y₂ - y₁)²]

Here (x₁, y₁)  =  G(1, 2) and (x₂, y₂)  =  H(6, 5)

GH  =  √[(6 - 1)² + (5 - 2)²]

GH  =  √[5² + 3²]

GH  =  √[25 + 9]

GH  =  √34

Conclusion :

Because BC = √34 and GH = √34,

BC  ≅  GH

All the three pairs of corresponding sides are congruent. By SSS congruence postulate,

ΔABC  ≅  ΔFGH

Problem 2 :

In the diagram given below, prove that ΔABC  ≅  ΔDEF

Solution :

From the diagram given above, we have

A(-3, 3), B(0, 1), C(-3, 1), D(0, 6), E(2, 3), F(2, 6)

Because AC = 2 in triangle ABC and DF = 2 in triangle DEF,

AC  ≅  DF.

Because BC = 3 in triangle ABC and EF = 3 in triangle DEF,

BC  ≅  EF.

Use the distance formula to find the lengths of BC and GH.

Length of AB :

AB  =  √[(x₂ - x₁)² + (y₂ - y₁)²]

Here (x₁, y₁)  =  A(-3, 3) and (x₂, y₂)  =  B(0, 1)

AB  =  √[(0 + 3)² + (1 - 3)²]

AB  =  √[3² + (-2)²]

AB  =  √[9 + 4]

AB  =  √13

Length of DE :

DE  =  √[(x₂ - x₁)² + (y₂ - y₁)²]

Here (x₁, y₁)  =  D(0, 6) and (x₂, y₂)  =  E(2, 3)

DE  =  √[(2 - 0)² + (3 - 6)²]

DE  =  √[2² + (-3)²]

DE  =  √[4 + 9]

DE  =  √13

Conclusion :

Because AB = √13 and DE = √13,

AB  ≅  DE

All the three pairs of corresponding sides are congruent. By SSS congruence postulate,

ΔABC  ≅  ΔDEF

Problem 3 :

In the diagram given below, prove that ΔOPM  ≅  ΔMNP

Solution :

PM is the common side for both the triangles OPM and MNP.

Because OP = 6 in triangle OPM and PN = 6 in triangle MNP,

OP  ≅  PN.

Use the distance formula to find the lengths of OM and MN.

Length of OM :

OM  =  √[(x₂ - x₁)² + (y₂ - y₁)²]

Here (x₁, y₁)  =  O(0, 0) and (x₂, y₂)  =  M(3, 3)

OM  =  √[(3 - 0)² + (3 - 0)²]

OM  =  √[3² + 3²]

OM  =  √[9 + 9]

OM  =  √18

Length of MN :

MN  =  √[(x₂ - x₁)² + (y₂ - y₁)²]

Here (x₁, y₁)  =  M(3, 3) and (x₂, y₂)  =  N(6, 6)

MN  =  √[(6 - 3)² + (6 - 3)²]

MN  =  √[3² + 3²]

MN  =  √[9 + 9]

MN  =  √18

Conclusion :

Because  OM = √18 and MN = √18,

OM  ≅  MN

All the three pairs of corresponding sides are congruent. By SSS congruence postulate,

ΔOPM  ≅  ΔMNP

After having gone through the stuff given above, we hope that the students would have understood how to do problems on congruent triangles on a coordinate plane.

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