Parallelogram Worksheet Answers - Fill Online, Printable, Fillable ... - Free Printable

Since I can't view or access uploaded images directly, I can’t see the specific image you're referring to. However, based on your description — a "Parallelogram Worksheet" with problems involving properties of parallelograms — I can help you solve and explain typical problems like these.

Let me walk through the common types of problems found in such worksheets and provide detailed explanations. If you can describe the exact questions (e.g., "In parallelogram ABCD, if AB = 8, what is CD?" or "If angle A is 70°, what is angle C?"), I can tailor my response accordingly.

But here’s a general guide for solving parallelogram-related problems, using standard geometry principles:

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🔷 Key Properties of Parallelograms:

1. Opposite sides are congruent
→ $ AB = CD $, $ AD = BC $

2. Opposite angles are congruent
→ $ \angle A = \angle C $, $ \angle B = \angle D $

3. Consecutive angles are supplementary
→ $ \angle A + \angle B = 180^\circ $, etc.

4. Diagonals bisect each other
→ The diagonals intersect at their midpoints.

5. Each diagonal divides the parallelogram into two congruent triangles

6. If one angle is known, others can be found using above rules

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Let's Solve Example Problems from Your Worksheet

Assuming the worksheet has problems like these:

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#### I. Complete each statement:

1. In a parallelogram, opposite sides are congruent.
2. In a parallelogram, opposite angles are congruent.
3. In a parallelogram, adjacent angles are supplementary to each other, which means they add up to 180°.

✔ These are definitions.

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#### II. Complete each statement, using Parallelogram ABCD

Let’s assume a standard diagram where:
- Points are labeled clockwise: A, B, C, D
- Diagonals AC and BD intersect at point E

Now solve:

4. If $ AB = 13 $, then $ CD = 13 $
→ Opposite sides are equal.

5. If $ \angle B = 22^\circ $, then $ \angle D = 22^\circ $
→ Opposite angles are equal.

6. If $ \angle A = 18^\circ $, then $ \angle C = 18^\circ $
→ Opposite angles.

7. If $ m\angle ABC = 110^\circ $, then $ m\angle ADC = 110^\circ $
→ Opposite angles.

8. If $ m\angle DAB = 75^\circ $, then $ m\angle BCD = 75^\circ $
→ Opposite angles.

9. If $ m\angle ABC = 75^\circ $, then $ m\angle BCD = ? $
→ Consecutive angles are supplementary:
$ m\angle BCD = 180^\circ - 75^\circ = 105^\circ $

10. If $ m\angle A = 38^\circ $, then $ m\angle D = ? $
→ $ \angle A $ and $ \angle D $ are consecutive → sum to 180°
So $ m\angle D = 180^\circ - 38^\circ = 142^\circ $

11. If $ m\angle ABC = 130^\circ $, then $ m\angle C = ? $
→ $ \angle C $ is opposite $ \angle A $, but we don’t have $ \angle A $. Wait — $ \angle ABC $ is $ \angle B $, so $ \angle D = 130^\circ $. Then $ \angle A = 180^\circ - 130^\circ = 50^\circ $, so $ \angle C = 50^\circ $ (opposite $ \angle A $).

12. If $ EF = x^2 + 4 $, $ BF = 2x + 5 $, $ DF = 2x + 2 $, and $ AE = 14 $, find $ x $.
→ Since diagonals bisect each other:
$ AE = EC $, $ BE = ED $

We’re told $ AE = 14 $, so $ EC = 14 $, thus $ AC = 28 $

But we need more info. Let's suppose E is intersection point, and $ BE = ED $, $ AE = EC $

Given:
- $ EF = x^2 + 4 $ → probably typo; maybe it's $ BE = x^2 + 4 $?
Wait — likely: $ BE = x^2 + 4 $, $ BF = 2x + 5 $, $ DF = 2x + 2 $

But if $ E $ is midpoint of diagonals, then $ BE = ED $

So $ BE = ED $

But $ BF $ and $ DF $ suggest F might be another point... unless F is E?

Possibility: typo, and “F” should be “E”

Assume:
- $ BE = x^2 + 4 $
- $ DE = 2x + 2 $
And since diagonals bisect each other: $ BE = DE $

So:
$$
x^2 + 4 = 2x + 2 \\
x^2 - 2x + 2 = 0
$$

Discriminant: $ (-2)^2 - 4(1)(2) = 4 - 8 = -4 $ → no real solution

So maybe different interpretation.

Alternatively, suppose:
- $ BF = 2x + 5 $, $ DF = 2x + 2 $
- And $ BE = ED $, so $ E $ is midpoint

But without knowing where F is, this is ambiguous.

Wait — perhaps F is the same as E?

Maybe: $ BE = x^2 + 4 $, $ DE = 2x + 2 $, and $ BE = DE $ → again:

$$
x^2 + 4 = 2x + 2 \\
x^2 - 2x + 2 = 0 \quad \text{No real solution}
$$

So maybe the problem says:
$ EF = x^2 + 4 $, $ BF = 2x + 5 $, $ DF = 2x + 2 $, and $ AE = 14 $

Still unclear.

Wait — maybe E is the intersection point, and F is a typo.

Another possibility: AE = 14, and BE = x² + 4, EC = 2x + 5, and since diagonals bisect each other:
$ AE = EC $ → $ 14 = 2x + 5 $ → $ 2x = 9 $ → $ x = 4.5 $

Then $ BE = x^2 + 4 = (4.5)^2 + 4 = 20.25 + 4 = 24.25 $

And $ DE = BE = 24.25 $

So if that’s the setup, then $ x = 4.5 $

But let’s confirm:
If $ AE = EC $, and $ AE = 14 $, then $ EC = 14 $, so $ 2x + 5 = 14 $ → $ x = 4.5 $

Yes.

So answer: $ x = 4.5 $ or $ \frac{9}{2} $

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#### III. Find the missing measurements of Parallelogram ABCD

Let’s assume standard labeling: A–B
| |
D–C

With diagonals AC and BD intersecting at E.

13. $ CD = ? $ → If $ AB = 10 $, then $ CD = 10 $ (opposite side)

14. $ DA = ? $ → If $ BC = 12 $, then $ DA = 12 $

15. $ AC = ? $ → If $ AE = 6 $, then $ AC = 2 \times 6 = 12 $

16. $ DB = ? $ → If $ DE = 8 $, then $ DB = 16 $

17. $ m\angle A = ? $ → If $ m\angle B = 110^\circ $, then $ m\angle A = 70^\circ $ (supplementary)

18. $ m\angle C = ? $ → Same as $ \angle A $, so $ 70^\circ $

19. $ m\angle BCD = ? $ → $ \angle C $, so $ 70^\circ $

20. $ m\angle ADC = ? $ → $ \angle D $, opposite $ \angle B $, so $ 110^\circ $

21. $ m\angle BAC = ? $ → This requires triangle info. Suppose $ \triangle ABC $, and $ \angle ABC = 110^\circ $, $ AB = BC $? Not necessarily.

But if $ AB \parallel CD $, and $ AC $ is diagonal, we need more.

Without specific values, hard to say.

But often: if $ \angle A = 70^\circ $, and $ \angle BAC $ is part of it, and triangle is isosceles? Not guaranteed.

So unless given more, can't determine.

22. $ m\angle BCA = ? $ → Again, depends on triangle.

But if $ \angle A = 70^\circ $, and $ \angle B = 110^\circ $, then in $ \triangle ABC $, $ \angle BAC + \angle BCA + \angle ABC = 180^\circ $

But $ \angle ABC = 110^\circ $, so $ \angle BAC + \angle BCA = 70^\circ $

But unless AB = BC, we can’t split.

So not enough info.

Wait — perhaps the diagram shows something.

Common trick: if diagonals are drawn, and angles are marked.

For example:

23. $ m\angle AED = ? $ → Vertical angles or triangle angles.

Suppose $ \angle AEB = 120^\circ $, then $ \angle AED = 60^\circ $ (linear pair)

Or if $ \angle AEB = 120^\circ $, then $ \angle AED = 60^\circ $, since straight line.

24. $ m\angle EBC = ? $ → Half of $ \angle B $? Only if diagonal bisects angle — not true in general.

Only in rhombus.

So unless specified, cannot assume.

25. $ m\angle BAE = ? $ → Again, depends.

26. $ m\angle CED = ? $ → Vertical to $ \angle AEB $, so same measure.

27. $ m\angle DBC = ? $ → Similar logic.

28. $ m\angle ADB = ? $ → Part of triangle.

29. $ m\angle EAB = ? $ → Same as $ \angle BAC $

30. $ m\angle AEB = ? $ → Often given or solvable via triangle sum.

31. $ m\angle BDC = ? $ → Opposite angle to $ \angle ABD $? No.

But $ \angle BDC $ is in triangle BDC.

If $ \angle CBD = x $, $ \angle BCD = y $, then $ \angle BDC = 180 - x - y $

32. $ m\angle CEB = ? $ → Vertical to $ \angle AED $

33. $ m\angle CEA = ? $ → Linear pair with $ \angle AEB $

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✔ Summary of How to Approach These Problems:

1. Use properties of parallelograms (opposite sides/angles equal, adjacent angles supplementary, diagonals bisect each other).
2. Label the diagram carefully.
3. Use algebra when variables are involved (like in #12).
4. Look for vertical angles, linear pairs, triangle sums when diagonals are involved.
5. Remember: Diagonals bisect each other → segments from center to vertices are equal.

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🛠️ Example Problem Solved:

> In parallelogram ABCD, $ AB = 10 $, $ \angle A = 60^\circ $. Find $ CD $, $ \angle C $, $ \angle B $, and $ \angle D $.

Solution:
- $ CD = AB = 10 $ (opposite sides)
- $ \angle C = \angle A = 60^\circ $ (opposite angles)
- $ \angle B = 180^\circ - 60^\circ = 120^\circ $ (adjacent angles)
- $ \angle D = \angle B = 120^\circ $

✔ Answer: $ CD = 10 $, $ \angle C = 60^\circ $, $ \angle B = 120^\circ $, $ \angle D = 120^\circ $

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📌 Final Note:

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