Transformations And Congruence Worksheet Answer Key

Transformations and Congruence Worksheet Answer Key

Introduction

In geometry, transformations such as translations, rotations, reflections, and dilations are the building blocks for understanding shape relationships. When students complete worksheets that ask them to identify congruent figures or apply specific transformations, they need a reliable reference to check their work. This answer key provides clear, step‑by‑step solutions for a typical Transformations and Congruence Worksheet. It is designed to help students verify their answers, understand the reasoning behind each step, and reinforce key concepts such as congruence criteria, image–preimage relationships, and transformational properties.

1. Translation Problems

Problem 1

Question: Translate triangle ( \triangle ABC ) by vector ( \langle 3, -2 \rangle ). What are the coordinates of the image triangle ( \triangle A'B'C' )?

Answer:
Add the vector components to each vertex:

• ( A(1, 4) \rightarrow A'(1+3, 4-2) = (4, 2) )
• ( B(5, 6) \rightarrow B'(5+3, 6-2) = (8, 4) )
• ( C(2, 9) \rightarrow C'(2+3, 9-2) = (5, 7) )

Result: ( \triangle A'B'C' ) has vertices ( (4, 2), (8, 4), (5, 7) ).

Problem 2

Question: A square has vertices at ( (0,0), (0,3), (3,3), (3,0) ). Translate it by vector ( \langle -2, 5 \rangle ). List the new vertices It's one of those things that adds up..

Answer:
Apply the vector to each point:

• ( (0,0) \rightarrow (-2, 5) )
• ( (0,3) \rightarrow (-2, 8) )
• ( (3,3) \rightarrow (1, 8) )
• ( (3,0) \rightarrow (1, 5) )

Result: New vertices are ( (-2, 5), (-2, 8), (1, 8), (1, 5) ) Which is the point..

2. Rotation Problems

Problem 3

Question: Rotate point ( P(4, 1) ) (90^\circ) counterclockwise around the origin.

Answer:
For a (90^\circ) CCW rotation, the transformation matrix is: [ \begin{pmatrix} 0 & -1\ 1 & 0 \end{pmatrix} ] Multiply: [ \begin{pmatrix} 0 & -1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 4\ 1 \end{pmatrix}

\begin{pmatrix} -1\ 4 \end{pmatrix} ] Result: ( P'(-1, 4) ).

Problem 4

Question: Rotate the triangle with vertices ( A(2,0), B(4,2), C(0,3) ) by (180^\circ) around point ( (1,1) ). Find the image triangle ( A'B'C' ).

Answer:
A (180^\circ) rotation around a point ( (h,k) ) can be performed by reflecting the point across ( (h,k) ) twice, or simply using the formula: [ (x', y') = (2h - x, 2k - y) ]

Apply to each vertex:

• ( A(2,0) \rightarrow A'(21-2, 21-0) = (0, 2) )
• ( B(4,2) \rightarrow B'(21-4, 21-2) = (-2, 0) )
• ( C(0,3) \rightarrow C'(21-0, 21-3) = (2, -1) )

Result: ( \triangle A'B'C' ) has vertices ( (0,2), (-2,0), (2,-1) ) Worth keeping that in mind..

3. Reflection Problems

Problem 5

Question: Reflect point ( Q(3, 5) ) over the ( y )-axis.

Answer:
Reflection over the ( y )-axis changes the sign of the ( x )-coordinate: [ (x', y') = (-x, y) ] So, [ Q'( -3, 5 ) ]

Problem 6

Question: Reflect the rectangle with vertices ( (1,2), (4,2), (4,5), (1,5) ) over the line ( y = x ) Nothing fancy..

Answer:
Reflection over ( y = x ) swaps the ( x ) and ( y ) coordinates:

• ( (1,2) \rightarrow (2,1) )
• ( (4,2) \rightarrow (2,4) )
• ( (4,5) \rightarrow (5,4) )
• ( (1,5) \rightarrow (5,1) )

Result: New vertices are ( (2,1), (2,4), (5,4), (5,1) ) That alone is useful..

4. Dilation Problems

Problem 7

Question: Dilate triangle ( \triangle DEF ) with center at the origin and scale factor ( k = 2 ). Original vertices: ( D(1,2), E(3,4), F(0,5) ). Find the dilated vertices Most people skip this — try not to..

Answer:
Multiply each coordinate by the scale factor ( k ):

• ( D(1,2) \rightarrow D'(2,4) )
• ( E(3,4) \rightarrow E'(6,8) )
• ( F(0,5) \rightarrow F'(0,10) )

Result: ( \triangle D'E'F' ) vertices are ( (2,4), (6,8), (0,10) ).

Problem 8

Question: Dilate a square with vertices ( (2,2), (5,2), (5,5), (2,5) ) about point ( (3,3) ) with a factor of ( \frac{1}{2} ). Determine the new vertices Worth keeping that in mind..

Answer:
Use the dilation formula relative to center ( (h,k) ): [ (x', y') = (h + k(x-h),; k + k(y-k)) ] Here ( h = 3, k = 3, ) and scaling factor ( s = \frac{1}{2} ).

Compute each vertex:

1. ( (2,2) \rightarrow (3 + \tfrac12(2-3), 3 + \tfrac12(2-3)) = (3 - 0.5, 3 - 0.5) = (2.5, 2.5) )
2. ( (5,2) \rightarrow (3 + \tfrac12(5-3), 3 + \tfrac12(2-3)) = (3 + 1, 3 - 0.5) = (4, 2.5) )
3. ( (5,5) \rightarrow (3 + \tfrac12(5-3), 3 + \tfrac12(5-3)) = (4, 4) )
4. ( (2,5) \rightarrow (3 + \tfrac12(2-3), 3 + \tfrac12(5-3)) = (2.5, 4) )

Result: The dilated square has vertices ( (2.5, 2.5), (4, 2.5), (4, 4), (2.5, 4) ) And that's really what it comes down to..

5. Congruence Problems

Problem 9

Question: Determine whether triangles ( \triangle ABC ) with sides ( 3, 4, 5 ) and ( \triangle DEF ) with sides ( 3, 4, 5 ) are congruent. Explain your reasoning.

Answer:
Both triangles satisfy the Side‑Side‑Side (SSS) Congruence Criterion: all three corresponding sides are equal. So, ( \triangle ABC \cong \triangle DEF ). The triangles are mirror images or rotated versions of each other, but congruence only requires side equality, not orientation That's the part that actually makes a difference..

Problem 10

Question: Triangle ( \triangle GHI ) has angles ( 30^\circ, 60^\circ, 90^\circ ). Triangle ( \triangle JKL ) has angles ( 60^\circ, 30^\circ, 90^\circ ). Are the triangles congruent? Justify.

Answer:
Triangles are congruent if all corresponding angles and sides match. Here, the angle sets are the same (just reordered). Since the side lengths opposite equal angles in a right triangle with those angles are in a fixed ratio ( 1:\sqrt{3}:2 ), the triangles are congruent by the Angle‑Angle‑Side (AAS) Criterion (one side and the two non‑adjacent angles are equal). Thus, ( \triangle GHI \cong \triangle JKL ).

Problem 11

Question: Verify if the following two triangles are congruent:

• Triangle ( \triangle MNO ) with side lengths ( 7, 24, 25 ).
• Triangle ( \triangle PQR ) with side lengths ( 7, 25, 24 ).

Provide a step‑by‑step explanation.

Answer:

1. Check SSS Criterion: All three sides of ( \triangle MNO ) are ( 7, 24, 25 ).
   For ( \triangle PQR ), the sides are ( 7, 25, 24 ).
2. Reorder: The sets are identical when reordered: ( {7, 24, 25} ).
3. Conclusion: Since every side of one triangle matches a side of the other, the triangles satisfy the SSS criterion.
   Result: ( \triangle MNO \cong \triangle PQR ).

6. Combined Transformation and Congruence

Problem 12

Question: A triangle ( \triangle ABC ) with vertices ( (0,0), (4,0), (0,3) ) is first reflected over the line ( y = x ) and then rotated (90^\circ) clockwise around the origin. Determine the final coordinates of the image triangle and confirm whether it is congruent to the original.

Answer:
Step 1 – Reflection over ( y=x ): Swap ( x ) and ( y ) for each vertex Worth keeping that in mind..

• ( A(0,0) \rightarrow A'(0,0) )
• ( B(4,0) \rightarrow B'(0,4) )
• ( C(0,3) \rightarrow C'(3,0) )

Step 2 – Rotation (90^\circ) clockwise: Use the matrix [ \begin{pmatrix} 0 & 1\ -1 & 0 \end{pmatrix} ] Apply to each reflected vertex:

• ( A'(0,0) \rightarrow A''(0,0) )
• ( B'(0,4) \rightarrow B''(4,0) )
• ( C'(3,0) \rightarrow C''(0,-3) )

Final vertices: ( (0,0), (4,0), (0,-3) ).

Congruence Check:

• Original triangle sides: ( AB = 4, AC = 3, BC = 5 ).
• Image triangle sides: ( A''B'' = 4, A''C'' = 3, B''C'' = 5 ).
  All side lengths match; angles are preserved by rigid motions. Thus, the image triangle is congruent to the original.

7. FAQ – Common Pitfalls

Conclusion

A solid grasp of transformations—translations, rotations, reflections, and dilations—paired with the proper use of congruence criteria, equips students to solve complex geometry problems confidently. This answer key not only provides the correct solutions but also illustrates the reasoning behind each step, reinforcing the logical flow essential for mastering geometry. Use it as a study aid, a quick reference, or a teaching tool to help learners internalize the principles that make shapes behave the way they do under transformation Most people skip this — try not to..