Congruence for Quadrilaterals

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Created by Illustrative MathematicsIllustrative Mathematics, CC BY 4.0
Subject: Geometry
Grade: 9-12
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23 questions

Congruence for Quadrilaterals

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Problem 1

1) Select all\textbf{all} quadrilaterals that have 180 degree rotational symmetry. Write each corresponding letter in the answer box and separate letters with commas.

a) trapezoid \quad\quad b) isosceles trapezoid \quad\quad c) parallelogram \quad\quad d) rhombus \quad\quad e) rectangle \quad\quad f) square

Problem 2

Lin wrote a proof to show that diagonal EGEG is a line of symmetry for rhombus EFGHEFGH. Fill in the blanks to complete her proof:

Because EFGHEFGH is a rhombus, the distance from EE to 1\underline{\quad\quad 1 \quad\quad} is the same as the distance from EE to 2\underline{\quad\quad 2 \quad\quad}. Since EE is the same distance from 3\underline{\quad\quad 3 \quad\quad} as it is from 4\underline{\quad\quad 4 \quad\quad}, it must lie on the perpendicular bisector of segment 5\underline{\quad\quad 5 \quad\quad}. By the same reasoning, GG must lie on the perpendicular bisector of 6\underline{\quad\quad 6 \quad\quad}. Therefore, line 7\underline{\quad\quad 7 \quad\quad} is the perpendicular bisector of segment FHFH. So reflecting rhombus EFGHEFGH across line 8\underline{\quad\quad 8 \quad\quad} will take EE to 9\underline{\quad\quad 9 \quad\quad} and GG to 10\underline{\quad\quad 10 \quad\quad} (because EE and GG are on the line of reflection) and FF to 11\underline{\quad\quad 11 \quad\quad} and HH to 12\underline{\quad\quad 12 \quad\quad} (since FHFH is perpendicular to the line of reflection, and FF and HH are the same distance from the line of reflection, on opposite sides). Since the image of rhombus EFGHEFGH reflected across EGEG is rhombus EHGFEHGF (the same rhombus!), line EFEF must be a line of symmetry for rhombus EFGHEFGH.

Fill in the blanks using items from the Bank of Terms below. Some items might be used more than once.

Bank of Terms: EE, FF, GG, HH, EGEG, FHFH

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2) Blanks 1 and 2, separated by a comma

3) Blanks 3 and 4, separated by a comma

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Problem 3

In quadrilateral ABCDABCD, ADAD is congruent to BCBC, and ADAD is parallel to BCBC. Andre has written a proof to show that ABCDABCD is a parallelogram. Fill in the blanks to complete the proof.

Since ADAD is parallel to 1\underline{\quad\quad 1 \quad\quad}, alternate interior angles 2\underline{\quad\quad 2 \quad\quad} and 3\underline{\quad\quad 3 \quad\quad} are congruent. ACAC is congruent to 4\underline{\quad\quad 4 \quad\quad} since segments are congruent to themselves. Along with the given information that ADAD is congruent to BCBC, triangle ADCADC is congruent to5\underline{\quad\quad 5 \quad\quad} by the 6\underline{\quad\quad 6 \quad\quad} Triangle Congruence. Since the triangles are congruent, all pairs of corresponding angles are congruent, so angle DCADCA is congruent to 7\underline{\quad\quad 7 \quad\quad}. Since those alternate interior angles are congruent, ABAB must be parallel to 8\underline{\quad\quad 8 \quad\quad}. Since we define a parallelogram as a quadrilateral with both pairs of opposite sides parallel, ABCDABCD is a parallelogram.

Fill in the blanks using items from the Bank of Terms below. Some items might be used more than once.

Bank of Terms: ACAC, BCBC, CDCD, BACBAC, BCABCA, CBACBA, DACDAC, Side-Angle-Side

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12) Blank 1

13) Blanks 2 and 3, separated by a comma

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Problem 4

19) Select the statement that must\textbf{must} be true.

a)

Parallelograms have at least one right angle.

b)

If a quadrilateral has opposite sides that are both congruent and parallel, then it is a parallelogram.

c)

Parallelograms have congruent diagonals.

d)

The height of a parallelogram is greater than the lengths of the sides.

e)


f)


Problem 5

20) EFGHEFGH is a parallelogram and angle HEFHEF is a right angle. Select all\textbf{all} statements that must\textbf{must} be true. Write each corresponding letter in the answer box and separate letters with commas.

a) EFGHEFGH is a rectangle. \quad\quad b) Triangle HEFHEF is congruent to triangle GFHGFH. \quad\quad c) Triangle HEFHEF is congruent to triangle FGHFGH. \quad\quad d) EDED is congruent to HDHD, DGDG, and DFDF. \quad\quad e) Triangle EDHEDH is congruent to triangle HDGHDG.

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Problem 6

Figure ABCDABCD is a parallelogram.

Given: ABCD\overline{AB} \cong \overline{CD}, ADBCBD\angle ADB \cong \angle CBD

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21) Is triangle ADBADB congruent to triangle CBDCBD?

True or false? Write below.

22) Explain your reasoning.

Problem 7

23) Figure KLMNKLMN is a parallelogram. Prove that triangle KNLKNL is congruent to triangle MLNMLN.

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