Math 1 Quiz: Rigid Transformations
2 questions · exam conditions
0:00
Rigid TransformationsQuestion 1 of 2

A regular hexagon centered at the origin undergoes a rotation of 240°240° counterclockwise about the origin. One vertex of the original hexagon is located at (4,0)(4, 0). After the rotation, this vertex has coordinates (a,b)(a, b). What is the value of a+ba + b?

−2−23-2 - 2\sqrt{3}
2−232 - 2\sqrt{3}
−2+23-2 + 2\sqrt{3}
2+232 + 2\sqrt{3}
← Back to quizzes

Math 1 Quiz

Math 1 Quiz: Rigid Transformations

Practice Rigid Transformations in Math 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rigid Transformations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A regular hexagon centered at the origin undergoes a rotation of 240°240° counterclockwise about the origin. One vertex of the original hexagon is located at (4,0)(4, 0). After the rotation, this vertex has coordinates (a,b)(a, b). What is the value of a+ba + b?

  1. −2−23-2 - 2\sqrt{3} (correct answer)
  2. 2−232 - 2\sqrt{3}
  3. −2+23-2 + 2\sqrt{3}
  4. 2+232 + 2\sqrt{3}
Explanation: For a rotation of 240°240° counterclockwise about the origin, the transformation is (x,y)→(xcos⁡240°−ysin⁡240°,xsin⁡240°+ycos⁡240°)(x, y) \rightarrow (x\cos 240° - y\sin 240°, x\sin 240° + y\cos 240°). We have cos⁡240°=cos⁡(180°+60°)=−cos⁡60°=−12\cos 240° = \cos(180° + 60°) = -\cos 60° = -\frac{1}{2} and sin⁡240°=sin⁡(180°+60°)=−sin⁡60°=−32\sin 240° = \sin(180° + 60°) = -\sin 60° = -\frac{\sqrt{3}}{2}. Applying this to the point (4,0)(4, 0): x′=4⋅(−12)−0⋅(−32)=−2x' = 4 \cdot \left(-\frac{1}{2}\right) - 0 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -2. y′=4⋅(−32)+0⋅(−12)=−23y' = 4 \cdot \left(-\frac{\sqrt{3}}{2}\right) + 0 \cdot \left(-\frac{1}{2}\right) = -2\sqrt{3}. Therefore, (a,b)=(−2,−23)(a, b) = (-2, -2\sqrt{3}) and a+b=−2+(−23)=−2−23a + b = -2 + (-2\sqrt{3}) = -2 - 2\sqrt{3}. The other choices represent common sign errors or incorrect angle calculations (like using 120°120° instead of 240°240°, or confusing sine and cosine values).

Question 2

Parallelogram MNOPMNOP has vertices M(1,4)M(1, 4), N(5,6)N(5, 6), O(7,2)O(7, 2), and P(3,0)P(3, 0). The parallelogram undergoes transformation T1T_1 (reflection across the xx-axis) followed by transformation T2T_2 (rotation 90°90° counterclockwise about the origin). What is the area of the image parallelogram after both transformations?

  1. 1212 square units
  2. 1616 square units
  3. 2020 square units (correct answer)
  4. 2424 square units
Explanation: Since rigid transformations preserve area, the area of the image parallelogram equals the area of the original parallelogram MNOPMNOP. To find the area, I'll use the cross product formula. First, find vectors MN→=(5−1,6−4)=(4,2)\overrightarrow{MN} = (5-1, 6-4) = (4, 2) and MP→=(3−1,0−4)=(2,−4)\overrightarrow{MP} = (3-1, 0-4) = (2, -4). The area is ∣MN→×MP→∣=∣4⋅(−4)−2⋅2∣=∣−16−4∣=∣−20∣=20|\overrightarrow{MN} \times \overrightarrow{MP}| = |4 \cdot (-4) - 2 \cdot 2| = |-16 - 4| = |-20| = 20 square units. We can verify this is a parallelogram since opposite sides are equal: MN→=PO→=(4,2)\overrightarrow{MN} = \overrightarrow{PO} = (4, 2) and MP→=NO→=(2,−4)\overrightarrow{MP} = \overrightarrow{NO} = (2, -4).