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Topic 2 - Parallel and Perpendicular Lines EOC Review

Authored by cari buddman

Mathematics

9th - 12th Grade

CCSS covered

Used 5+ times

Topic 2 - Parallel and Perpendicular Lines EOC Review
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10 questions

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1.

FILL IN THE BLANK QUESTION

2 mins • 1 pt

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What is m∠2 + m∠3?

(Just type the numerical answer, do not include degrees.)

Tags

CCSS.7.G.B.5

2.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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If m∠4 = 35, find m∠2. Explain.

55; ∠2 and ∠4 are complementary angles, so m∠2 = 90 - m∠4.

35; ∠2 and ∠4 are corresponding angles, so m∠2 = m∠4.

145; ∠2 and ∠4 are supplementary angles, so m∠2 = 180 - m∠4.

35; ∠2 and ∠4 are alternate interior angles, so m∠2 = m∠4.

Tags

CCSS.8.G.A.5

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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If m∠4 = 35, find m∠3. Explain.

55; ∠3 and ∠4 are complementary angles, so m∠3 = 90 - m∠4.

35; ∠3 and ∠4 are corresponding angles, so m∠3 = m∠4.

145; ∠3 and ∠4 are supplementary angles, so m∠3 = 180 - m∠4.

35; ∠3 and ∠4 are alternate interior angles, so m∠3 = m∠4.

Tags

CCSS.8.G.A.5

4.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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If ∠1 ≅ ∠2, can you conclude that any of the lines are parallel? Explain.

Yes; lines n and p are parallel because corresponding angles are congruent.

No; ∠1 and ∠2 show no relationship.

Yes; lines ℓ and m are parallel because corresponding angles are congruent.

No; neither angle is formed by the transversal, line q.

Tags

CCSS.8.G.A.5

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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If m∠3 + m∠4 = 180, which lines can you conclude are parallel? Explain.

Lines n and p are parallel because alternate interior angles are congruent.

Lines n and p are parallel because same-side interior angles are supplementary.

Lines ℓ and m are parallel because same-side interior angles are supplementary.

Lines ℓ and m are parallel because alternate interior angles are congruent.

Tags

CCSS.8.G.A.5

6.

FILL IN THE BLANK QUESTION

2 mins • 1 pt

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What is m∠1?

Tags

CCSS.8.G.A.5

7.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

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Which of the following is true?

m∠BCD − m∠DBC = 10°

m∠BCD − m∠DBC = 20°

m∠BCD − m∠DBC = 60°

m∠BCD − m∠DBC = 80°

Tags

CCSS.8.G.A.5

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