Geometry Worksheet 6.2 Parallelograms PDF | PDF - Free Printable

Let's solve this Geometry Worksheet on Parallelograms step by step, using the properties of parallelograms.

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Word Bank:

- parallelogram
- congruent
- supplementary
- bisect

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1–5. Fill in the blanks

1. A quadrilateral with two pairs of parallel sides is a parallelogram.
✔ *Definition of a parallelogram.*

2. The opposite sides of a parallelogram are congruent.
✔ *Property: Opposite sides are equal in length.*

3. The opposite angles of a parallelogram are congruent.
✔ *Opposite angles are equal.*

4. The consecutive angles of a parallelogram are supplementary.
✔ *Consecutive angles add up to 180°.*

5. The diagonals of a parallelogram bisect each other.
✔ *Diagonals cut each other into two equal parts.*

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6–8. Use parallelogram MATH

Given: Parallelogram MATH, with vertices labeled in order (M, A, T, H).

So, sides: MA || HT, MH || AT

Angles: ∠M, ∠A, ∠T, ∠H

6. Name two pairs of congruent angles.
- ∠M ≅ ∠T
- ∠A ≅ ∠H
✔ *Opposite angles are congruent.*

7. Name four pairs of supplementary angles.
Supplementary means they add to 180°. Consecutive angles are supplementary.

- ∠M and ∠A
- ∠A and ∠T
- ∠T and ∠H
- ∠H and ∠M

✔ These are adjacent (consecutive) angles.

8. Name two pairs of congruent segments.
- MA ≅ HT
- MH ≅ AT
✔ *Opposite sides are congruent.*

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9–12. Find missing angles and sides. Label on pictures.

#### Problem 9:
Parallelogram NOPQ
- Given:
- ON = 4
- OP = 6
- ∠N = 32°

We know:
- Opposite sides are congruent → OQ = 4, NP = 6
- Opposite angles are congruent → ∠P = 32°
- Consecutive angles are supplementary → ∠O = 180° − 32° = 148°
So ∠Q = 148°

✔ Label:
- ∠O = 148°
- ∠P = 32°
- ∠Q = 148°
- OQ = 4, NP = 6

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#### Problem 10:
Parallelogram WXYZ
- Given:
- WX = 7
- XY = 10
- ∠X = 138°

Then:
- Opposite sides: WZ = 10, YZ = 7
- Opposite angles: ∠Z = 138°
- Consecutive angles: ∠W = 180° − 138° = 42°
So ∠Y = 42°

✔ Label:
- ∠W = 42°
- ∠Y = 42°
- WZ = 10, YZ = 7

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#### Problem 11:
Parallelogram ABCD
- Given:
- AD = 13
- CD = 12
- ∠BAC = 24°
- ∠ACD = 15°

We need to find missing angles.

Note: Diagonal AC is drawn.

Let’s analyze triangle ABC and triangle ADC.

Since ABCD is a parallelogram:
- AB || CD
- AD || BC
- ∠ABC = ∠ADC
- ∠BAD = ∠BCD

But we’re given angles at diagonal AC.

In triangle ABC:
- ∠BAC = 24°
- We can use triangle angle sum or look for relationships.

Wait — notice that diagonal AC divides the parallelogram into two triangles.

Let’s focus on angles around point C:

∠ACD = 15°
But ∠ACD is part of ∠C.

Also, since AB || CD, and AC is transversal, then:

∠BAC and ∠ACD are alternate interior angles → they should be equal?

But here, ∠BAC = 24°, ∠ACD = 15° → not equal? That contradicts unless I misread.

Wait — maybe it's not alternate interior?

Let me draw mentally:

- AB || CD
- AC is diagonal → transversal

Then ∠BAC and ∠ACD are not alternate interior — they are on same side.

Alternate interior would be ∠BAC and ∠ACD only if the transversal cuts both lines — but ∠ACD is inside triangle.

Actually, better approach:

In parallelogram ABCD:
- AB || CD
- AD || BC

Now, consider triangle ABC and triangle CDA.

We are told:
- ∠BAC = 24°
- ∠ACD = 15°

But note: ∠ACD is the same as ∠ACB? No — wait.

Wait, point D is connected to C, so ∠ACD is angle at C between A-C-D.

So ∠ACD = 15° → this is part of angle C.

But angle C = ∠BCD = ∠BCA + ∠ACD?

No — actually, diagonal AC splits ∠C into two parts: ∠ACB and ∠ACD?

Wait — no! In parallelogram ABCD, the vertices are in order: A, B, C, D.

So diagonal AC connects A to C.

Then:
- At vertex C: angle ∠BCD is split by diagonal AC into ∠ACB and ∠ACD? Only if D is on one side.

Wait — actually, from point C, the sides are CB and CD.

So diagonal AC goes from A to C — so it splits angle C into ∠ACB and ∠ACD?

Yes.

So ∠C = ∠ACB + ∠ACD

But we are told ∠ACD = 15°

But we don’t know ∠ACB.

But also, since AB || CD, and AC is transversal:

Then ∠BAC and ∠ACD are alternate interior angles → so they should be equal.

But ∠BAC = 24°, ∠ACD = 15° → not equal → contradiction?

Wait — unless the labeling is different.

Wait — maybe I have the points wrong.

Standard labeling: ABCD in order → A to B to C to D.

So AB || CD, AD || BC.

Diagonal AC.

Now, line AB and line CD are parallel.

Transversal AC intersects them.

Then:
- ∠BAC (at A, between AB and AC)
- ∠ACD (at C, between CD and AC)

These are alternate interior angles → should be equal.

But here, ∠BAC = 24°, ∠ACD = 15° → not equal.

That suggests either the diagram is different or there's a mistake.

Wait — perhaps ∠ACD is not the angle at C between CD and AC?

But it is labeled as such.

Wait — maybe the angle given is not ∠ACD, but ∠ACB?

Wait — problem says: "∠ACD = 15°" — so at point C, between A-C-D.

But if AB || CD, then alternate interior angles: ∠BAC and ∠ACD must be equal.

So if ∠BAC = 24°, then ∠ACD must be 24°, not 15°.

But it says 15° → contradiction.

Unless the figure is not standard.

Wait — maybe the angle is not ∠ACD but ∠ACB?

Let me re-read: "15°" is at point C, labeled near C, probably ∠ACB?

Wait — the label is at point C, and it says "15°", with arc from AC to BC? Or from AC to DC?

Looking at the image description: "15°" is next to point C, likely between AC and DC → so ∠ACD = 15°.

But then conflict with alternate interior angles.

Alternatively, perhaps the angle at A is not ∠BAC, but something else.

Wait — maybe the 24° is ∠CAD?

But it says ∠BAC = 24°.

Wait — unless the diagonal is BD, not AC?

No — the diagonal is drawn from A to C.

Wait — perhaps I need to re-evaluate.

Alternative idea: Maybe the 24° is not ∠BAC, but ∠CAD?

But the problem says: "24°" at point A, between B-A-C → so ∠BAC = 24°.

And at point C, ∠ACD = 15°.

But in parallelogram, AB || CD → transversal AC → alternate interior angles: ∠BAC and ∠ACD should be equal.

So 24° ≠ 15° → impossible.

So likely, the 15° is not ∠ACD, but ∠ACB?

Wait — let's assume the diagram shows:

At point C, the angle between AC and BC is 15° → so ∠ACB = 15°

Then, since AB || CD, and AC is transversal, then ∠BAC and ∠ACD are alternate interior → so ∠ACD = ∠BAC = 24°

But then ∠C = ∠ACB + ∠ACD = 15° + 24° = 39°

But also, in parallelogram, consecutive angles are supplementary.

So ∠A = ∠BAC + ∠CAD?

Wait — we don't know ∠CAD.

Wait — we know ∠BAC = 24°, but what about ∠CAD?

Wait — diagonal AC divides ∠A into ∠BAC and ∠CAD.

But we don’t know ∠CAD.

But we do know side lengths: AD = 13, CD = 12, and AB = 13 (since opposite), BC = 12.

Wait — no: AD = 13, CD = 12 → so AB = 13, BC = 12

But we need more.

Wait — perhaps use triangle ABC and triangle ADC.

But without more info, maybe we can use the fact that in triangle ABC, we know:

- ∠BAC = 24°
- ∠ACB = 15° → assuming the 15° is ∠ACB

Then in triangle ABC:
- ∠ABC = 180° − 24° − 15° = 141°

Then, since ABCD is parallelogram:
- ∠D = ∠B = 141°
- ∠A = 180° − 141° = 39°
- But ∠A = ∠BAC + ∠CAD = 24° + ∠CAD → so ∠CAD = 15°

Then in triangle ADC:
- ∠DAC = 15°
- ∠ACD = ? → but earlier we said ∠ACD = 24° (from alternate interior), but now we have ∠ACB = 15°, so ∠C = 15° + ∠ACD = ?

Wait — if ∠C = 15° + ∠ACD, and ∠C = 141°? No — ∠C = ∠B = 141°? Wait no.

Earlier: ∠ABC = 141°, so ∠C = 180° − 141° = 39°

Yes!

So ∠C = 39°

And ∠C = ∠ACB + ∠ACD = 15° + ∠ACD → so ∠ACD = 24°

Now, check alternate interior angles: ∠BAC = 24°, ∠ACD = 24° → yes, equal → good.

So the 15° is ∠ACB, not ∠ACD.

So the labeling must be: at point C, the angle between AC and BC is 15° → ∠ACB = 15°

Then ∠ACD = 24°

And ∠C = 15° + 24° = 39°

Then ∠A = 180° − 39° = 141°

But ∠A = ∠BAC + ∠CAD = 24° + ∠CAD → so ∠CAD = 117°

But that seems large.

Wait — no: ∠A is the whole angle at A, which is ∠DAB.

In parallelogram, ∠A + ∠B = 180°

We found ∠B = 141° → so ∠A = 39°

But we have ∠BAC = 24° → so ∠CAD = 39° − 24° = 15°

So ∠CAD = 15°

Now in triangle ADC:
- ∠DAC = 15°
- ∠ACD = 24°
- So ∠ADC = 180° − 15° − 24° = 141°

But ∠ADC = ∠B = 141° → matches.

So everything checks.

So missing angles:

- ∠B = 141°
- ∠D = 141°
- ∠C = 39°
- ∠A = 39°

But ∠A is composed of ∠BAC = 24° and ∠CAD = 15° → total 39°

Now, sides:
- AB = 13 (since AD = 13, opposite)
- BC = 12 (CD = 12, opposite)
- All sides known.

So just label:
- ∠A = 39°, ∠B = 141°, ∠C = 39°, ∠D = 141°
- ∠CAD = 15°, ∠ACB = 15°, etc.

But the question asks to find missing angles and sides and label them.

So:
- ∠B = 141°
- ∠D = 141°
- ∠C = 39°
- ∠A = 39°
- ∠CAD = 15°
- ∠ACB = 15°
- ∠ACD = 24°
- Sides: AB = 13, BC = 12, CD = 12, DA = 13

✔ Label accordingly.

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#### Problem 12:
Parallelogram EFGH
- Given:
- ∠F = 70°
- ∠H = 20°
- Diagonals intersect at point X
- Markings: diagonals bisect each other (hash marks on diagonals)

So diagonals EG and FH intersect at X.

Hash marks:
- On EG: EX = XG → so diagonals bisect each other → expected
- On FH: FX = XH → same

Also, some hash marks on diagonals suggest equal segments.

But we are given:
- ∠F = 70° → ∠EFG = 70°
- ∠H = 20° → ∠GHE = 20°

But in parallelogram, opposite angles are equal.

So ∠F = ∠H → 70° = 20°? Contradiction.

Wait — unless ∠H is not ∠GHE, but ∠EHG?

Wait — the angle at H is ∠EHG.

But if ∠F = 70°, then ∠H = 70° (opposite angle)

But it says ∠H = 20° → conflict.

Wait — perhaps the 20° is not ∠H, but an angle in triangle?

Look: the diagram shows 20° at point H, between G-H and diagonal.

Wait — the angle marked 20° is likely ∠GHX or ∠CHX?

The marking shows 20° at H, between GH and diagonal.

So it's not the full angle at H.

Similarly, 70° at F, between EF and diagonal.

So likely:
- ∠EFH = 70° → angle between EF and diagonal FH
- ∠GHF = 20° → angle between GH and diagonal FH

Now, in parallelogram EFGH:
- EF || GH
- EH || FG

Diagonal FH connects F to H.

So in triangle EFH and triangle GHF.

But let’s use triangle properties.

At point F: ∠EFH = 70°

At point H: ∠GHF = 20°

Now, since EF || GH, and FH is transversal, then:

- ∠EFH and ∠GHF are alternate interior angles → should be equal?

But 70° ≠ 20° → contradiction.

Wait — no: alternate interior angles would be ∠EFH and ∠GHF only if they are on opposite sides.

Actually, ∠EFH and ∠GHF are on the same side of transversal FH → they are consecutive interior angles → should be supplementary?

But 70° + 20° = 90° ≠ 180° → not supplementary.

So again, conflict.

Wait — perhaps the 70° is ∠EFG, the full angle at F.

And the 20° is ∠GHF, part of angle at H.

But then opposite angles: ∠F = 70°, so ∠H = 70°

But ∠H is split by diagonal into ∠GHF and ∠EHF.

If ∠GHF = 20°, then ∠EHF = 50°

Similarly, at F: ∠EFG = 70°, split by diagonal into ∠EFH and ∠HFG.

But we don't know how it's split.

But we see that diagonals are marked with hash marks: on EG, EX = XG; on FH, FX = XH → so diagonals bisect each other.

Also, the diagonals are shown with tick marks: one on each half, so equal.

Now, the key: in parallelogram, diagonals bisect each other, but not necessarily equal.

Here, we are to find missing angles.

But we have:
- ∠F = 70° → so ∠H = 70°
- ∠E = ∠G = 110° (since consecutive angles are supplementary)

But we are given ∠H = 20°? No — the 20° is not the full angle.

Ah! The 20° is likely the angle between diagonal and side.

So at point H: ∠GHF = 20°

Then, since ∠H = 70°, the other part ∠EHF = 70° − 20° = 50°

Similarly, at F: ∠EFH = 70° → but that’s the full angle? Or part?

Wait — the 70° is labeled at F, between EF and FH → so ∠EFH = 70°

Then, since ∠F = 70°, and it's split by diagonal into ∠EFH and ∠HFG, then ∠HFG = 0°? Impossible.

Wait — unless the 70° is the full angle at F.

But then ∠EFH = 70°, but that would mean the diagonal FH coincides with FG — impossible.

So likely, the 70° is the full angle at F: ∠EFG = 70°

Then, diagonal FH splits it into ∠EFH and ∠HFG.

But we don't know how.

But at H, we have ∠GHF = 20°

Since EF || GH, and FH is transversal, then:

- ∠EFH and ∠GHF are alternate interior angles → so they should be equal.

So ∠EFH = ∠GHF = 20°

But we are told ∠EFH = 70° → conflict.

Wait — unless the 70° is not ∠EFH.

Let me re-express.

Perhaps the 70° is ∠EFG, the full angle at F.

Then, diagonal FH splits it into ∠EFH and ∠HFG.

Similarly, at H, diagonal FH splits ∠GHE into ∠GHF and ∠EHF.

Now, since EF || GH, and FH is transversal, then:

- ∠EFH and ∠GHF are alternate interior angles → so ∠EFH = ∠GHF

But we are given ∠GHF = 20° → so ∠EFH = 20°

Then, since ∠EFG = 70°, then ∠HFG = 70° − 20° = 50°

Similarly, at H: ∠GHE = 70° (since opposite to F), and ∠GHF = 20°, so ∠EHF = 50°

Now, in triangle EFH:
- ∠EFH = 20°
- ∠FEH = ?
- ∠EHF = 50°

Sum = 180° → ∠FEH = 180° − 20° − 50° = 110°

But ∠FEH is part of ∠E.

∠E = 110° (since consecutive to 70°), and ∠FEH = 110° → so the entire angle at E is ∠FEH, meaning no other part — so diagonal doesn't split it.

Similarly, at G: ∠G = 110°

Now, in triangle FGH:
- ∠GHF = 20°
- ∠HFG = 50°
- So ∠FGH = 180° − 20° − 50° = 110°

But ∠G = 110° → so again, entire angle.

So consistent.

So missing angles:
- ∠E = 110°
- ∠G = 110°
- ∠EFH = 20°
- ∠HFG = 50°
- ∠EHF = 50°
- ∠GHF = 20°

And diagonals bisect each other → so EX = XG, FX = XH

✔ Label all.

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13. Use parallelogram ABCD to answer:

Given:
- DX = 4
- AX = 6

Find:
- BX = ?
- BD = ?
- XC = ?
- AC = ?

Diagram: Parallelogram ABCD, diagonals AC and BD intersect at X.

In parallelogram, diagonals bisect each other.

So:
- AX = XC → since AC is bisected
- BX = XD → since BD is bisected

Given:
- DX = 4 → so BX = 4 (since diagonals bisect)
- Then BD = BX + XD = 4 + 4 = 8

- AX = 6 → so XC = 6
- AC = AX + XC = 6 + 6 = 12

✔ Answers:
- BX = 4
- BD = 8
- XC = 6
- AC = 12

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✔ Final Answers:

#### 1–5:
1. parallelogram
2. congruent
3. congruent
4. supplementary
5. bisect

#### 6–8:
6. ∠M ≅ ∠T and ∠A ≅ ∠H
7. ∠M and ∠A, ∠A and ∠T, ∠T and ∠H, ∠H and ∠M
8. MA ≅ HT and MH ≅ AT

#### 9–12: (as described above — label on diagrams)

#### 13:
- BX = 4
- BD = 8
- XC = 6
- AC = 12

Let me know if you'd like a visual version or explanation of any specific part!