**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. **

**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.

**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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