Geometry Worksheet 6.2 Parallelograms PDF | PDF - Free Printable
Educational worksheet: Geometry Worksheet 6.2 Parallelograms PDF | PDF. Download and print for classroom or home learning activities.
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Step-by-step solution for: Geometry Worksheet 6.2 Parallelograms PDF | PDF
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Show Answer Key & Explanations
Step-by-step solution for: Geometry Worksheet 6.2 Parallelograms PDF | PDF
Here are the solutions to the geometry worksheet problems, broken down step by step.
1. A quadrilateral with two pairs of parallel sides is a...
* Reasoning: This is the definition of the shape we are studying.
* Answer: parallelogram
2. The opposite sides of a parallelogram are...
* Reasoning: In a parallelogram, the top matches the bottom, and the left matches the right in length.
* Answer: congruent
3. The opposite angles of a parallelogram are...
* Reasoning: Angles across from each other (like top-left and bottom-right) are equal in measure.
* Answer: congruent
4. The consecutive angles of a parallelogram are...
* Reasoning: Consecutive angles are next to each other (neighbors). They add up to $180^\circ$.
* Answer: supplementary
5. The diagonals of a parallelogram... each other.
* Reasoning: When you draw lines connecting opposite corners, they cross exactly in the middle, cutting each other in half.
* Answer: bisect
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6. Name two pairs of congruent angles.
* Reasoning: Opposite angles are equal. Angle $M$ is opposite Angle $T$, and Angle $A$ is opposite Angle $H$.
* Answer: $\angle M \cong \angle T$ and $\angle A \cong \angle H$
7. Name four pairs of supplementary angles.
* Reasoning: Any two angles next to each other add up to $180^\circ$.
* Answer: $\angle M$ & $\angle A$, $\angle A$ & $\angle T$, $\angle T$ & $\angle H$, $\angle H$ & $\angle M$
8. Name two pairs of congruent segments.
* Reasoning: Opposite sides are equal in length. Side $MA$ is opposite side $HT$, and side $MT$ is opposite side $AH$.
* Answer: $MA \cong HT$ and $MT \cong AH$
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9. Parallelogram NOPQ
* Missing Side $PQ$: Opposite sides are equal. Side $NO = 4$, so side $PQ = 4$.
* Missing Side $OP$: Opposite sides are equal. Side $NQ = 6$, so side $OP = 6$.
* Angle $O$: Consecutive angles add to $180^\circ$. $180 - 32 = 148^\circ$.
* Angle $P$: Opposite angles are equal. Angle $N = 32^\circ$, so Angle $P = 32^\circ$.
* Angle $Q$: Opposite angles are equal. Angle $O = 148^\circ$, so Angle $Q = 148^\circ$.
10. Parallelogram WXYZ
* Side $YZ$: Opposite side $WX = 7$, so $YZ = 7$.
* Side $WZ$: Opposite side $XY = 10$, so $WZ = 10$.
* Angle $Y$: Opposite angle $W = 138^\circ$, so Angle $Y = 138^\circ$.
* Angle $X$: Consecutive to $W$. $180 - 138 = 42^\circ$.
* Angle $Z$: Opposite angle $X$, so Angle $Z = 42^\circ$.
11. Parallelogram ABCD
* Side $BC$: Opposite side $AD = 13$, so $BC = 13$.
* Side $AB$: Opposite side $CD = 12$, so $AB = 12$.
* Angle $ACB$: Alternate interior angles are equal because lines $AD$ and $BC$ are parallel. Angle $DAC = 24^\circ$, so Angle $ACB = 24^\circ$.
* Angle $BAC$: Alternate interior angles are equal because lines $AB$ and $CD$ are parallel. Angle $ACD = 15^\circ$, so Angle $BAC = 15^\circ$.
* Angle $D$: Look at triangle $ADC$. The angles must add to $180^\circ$. $180 - 24 - 15 = 141^\circ$.
* Angle $B$: Opposite Angle $D$, so Angle $B = 141^\circ$.
12. Rectangle EFGH
* *Note: The markings show this is a rectangle (diagonals are equal and bisect each other).*
* Side $EH$: Opposite side $FG = 5$, so $EH = 5$.
* Angle $GEH$: Corner angles of a rectangle are $90^\circ$. We know part of it ($\angle FEH$) is not given directly, but we can find parts using triangles. Let's look at the triangle formed by the diagonals.
* Actually, simpler approach: $\angle FEH = 90^\circ$. We are given $\angle FEG = 70^\circ$? No, the arc is for $\angle GEF$? Let's assume the label $70^\circ$ is $\angle GEF$ and $20^\circ$ is $\angle GEH$? Wait, looking closely at vertex E: The angle labeled $70^\circ$ is $\angle FEG$ (part of the corner). The angle labeled $20^\circ$ is at vertex H, specifically $\angle GHE$? No, it looks like $\angle EHG$ is split. Let's look at the diagonal intersection.
* Let's use alternate interior angles. Line $FG$ is parallel to $EH$. Diagonal $EG$ is a transversal. Therefore, $\angle FGE = \angle GEH$.
* Let's look at Vertex E. The corner is $90^\circ$. If the angle labeled $70^\circ$ is $\angle FEG$, then $\angle GEH = 90 - 70 = 20^\circ$.
* Let's check Vertex H. The corner is $90^\circ$. If the angle labeled $20^\circ$ is $\angle EHG$? No, usually these labels refer to the angle between the diagonal and the side. Let's assume the $20^\circ$ is $\angle GHE$? No, that would make triangle $EGH$ have angles $20, 90, 70$. That works.
* So, $\angle GEH = 20^\circ$ (complement of 70).
* $\angle FEG = 70^\circ$.
* $\angle EFG = 90^\circ$ and $\angle FGH = 90^\circ$ and $\angle GHE = 90^\circ$ and $\angle HEF = 90^\circ$.
* Using alternate interior angles: $\angle EGF = \angle GEH = 20^\circ$.
* $\angle FGE = 20^\circ$. Then in triangle $EFG$: $90 + 70 + 20 = 180$. Correct.
* $\angle HEG = 20^\circ$.
* $\angle EGH = 70^\circ$ (alternate interior to $\angle FEG$).
* $\angle FHG$? Diagonal $FH$. Triangle $FGH$ is congruent to $HEF$.
* Let's just list the obvious ones based on the diagram labels which seem to indicate:
* $\angle FEG = 70^\circ$
* $\angle GEH = 20^\circ$ (since $90-70=20$)
* $\angle EFG = 90^\circ, \angle FGH = 90^\circ, \angle GHE = 90^\circ, \angle HEF = 90^\circ$.
* By alternate interior angles: $\angle EGF = 20^\circ$ and $\angle FGE$? No, $\angle EGF$ is the angle at G inside triangle EFG. $\angle EGF = 20^\circ$.
* $\angle HGE = 70^\circ$.
* At the center intersection (let's call it P): Triangle $EPG$ has angles $20, 20, 140$. So vertical angle $\angle FPH = 140^\circ$. Adjacent angle $\angle EPF = 40^\circ$.
13. Parallelogram ABCD with Diagonals
* Given: $DX = 4$ and $AX = 6$.
* Rule: Diagonals bisect each other. This means $X$ is the midpoint of both $BD$ and $AC$.
* Find $BX$: Since $X$ is the midpoint of $BD$, $BX = DX$. So, $BX = 4$.
* Find $BD$: $BD = BX + DX = 4 + 4 = 8$.
* Find $XC$: Since $X$ is the midpoint of $AC$, $XC = AX$. So, $XC = 6$.
* Find $AC$: $AC = AX + XC = 6 + 6 = 12$.
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Final Answer:
1. parallelogram
2. congruent
3. congruent
4. supplementary
5. bisect
6. $\angle M \cong \angle T$ and $\angle A \cong \angle H$
7. $\angle M$ & $\angle A$, $\angle A$ & $\angle T$, $\angle T$ & $\angle H$, $\angle H$ & $\angle M$
8. $MA \cong HT$ and $MT \cong AH$
9. $PQ = 4$, $OP = 6$, $\angle O = 148^\circ$, $\angle P = 32^\circ$, $\angle Q = 148^\circ$
10. $YZ = 7$, $WZ = 10$, $\angle Y = 138^\circ$, $\angle X = 42^\circ$, $\angle Z = 42^\circ$
11. $BC = 13$, $AB = 12$, $\angle ACB = 24^\circ$, $\angle BAC = 15^\circ$, $\angle D = 141^\circ$, $\angle B = 141^\circ$
12. $EH = 5$, All corner angles are $90^\circ$. $\angle GEH = 20^\circ$, $\angle EGF = 20^\circ$, $\angle HGE = 70^\circ$.
13. $BX = 4$, $BD = 8$, $XC = 6$, $AC = 12$
Part 1: Word Bank (Questions 1–5)
1. A quadrilateral with two pairs of parallel sides is a...
* Reasoning: This is the definition of the shape we are studying.
* Answer: parallelogram
2. The opposite sides of a parallelogram are...
* Reasoning: In a parallelogram, the top matches the bottom, and the left matches the right in length.
* Answer: congruent
3. The opposite angles of a parallelogram are...
* Reasoning: Angles across from each other (like top-left and bottom-right) are equal in measure.
* Answer: congruent
4. The consecutive angles of a parallelogram are...
* Reasoning: Consecutive angles are next to each other (neighbors). They add up to $180^\circ$.
* Answer: supplementary
5. The diagonals of a parallelogram... each other.
* Reasoning: When you draw lines connecting opposite corners, they cross exactly in the middle, cutting each other in half.
* Answer: bisect
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Part 2: Parallelogram MATH (Questions 6–8)
6. Name two pairs of congruent angles.
* Reasoning: Opposite angles are equal. Angle $M$ is opposite Angle $T$, and Angle $A$ is opposite Angle $H$.
* Answer: $\angle M \cong \angle T$ and $\angle A \cong \angle H$
7. Name four pairs of supplementary angles.
* Reasoning: Any two angles next to each other add up to $180^\circ$.
* Answer: $\angle M$ & $\angle A$, $\angle A$ & $\angle T$, $\angle T$ & $\angle H$, $\angle H$ & $\angle M$
8. Name two pairs of congruent segments.
* Reasoning: Opposite sides are equal in length. Side $MA$ is opposite side $HT$, and side $MT$ is opposite side $AH$.
* Answer: $MA \cong HT$ and $MT \cong AH$
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Part 3: Find Missing Angles and Sides (Questions 9–12)
9. Parallelogram NOPQ
* Missing Side $PQ$: Opposite sides are equal. Side $NO = 4$, so side $PQ = 4$.
* Missing Side $OP$: Opposite sides are equal. Side $NQ = 6$, so side $OP = 6$.
* Angle $O$: Consecutive angles add to $180^\circ$. $180 - 32 = 148^\circ$.
* Angle $P$: Opposite angles are equal. Angle $N = 32^\circ$, so Angle $P = 32^\circ$.
* Angle $Q$: Opposite angles are equal. Angle $O = 148^\circ$, so Angle $Q = 148^\circ$.
10. Parallelogram WXYZ
* Side $YZ$: Opposite side $WX = 7$, so $YZ = 7$.
* Side $WZ$: Opposite side $XY = 10$, so $WZ = 10$.
* Angle $Y$: Opposite angle $W = 138^\circ$, so Angle $Y = 138^\circ$.
* Angle $X$: Consecutive to $W$. $180 - 138 = 42^\circ$.
* Angle $Z$: Opposite angle $X$, so Angle $Z = 42^\circ$.
11. Parallelogram ABCD
* Side $BC$: Opposite side $AD = 13$, so $BC = 13$.
* Side $AB$: Opposite side $CD = 12$, so $AB = 12$.
* Angle $ACB$: Alternate interior angles are equal because lines $AD$ and $BC$ are parallel. Angle $DAC = 24^\circ$, so Angle $ACB = 24^\circ$.
* Angle $BAC$: Alternate interior angles are equal because lines $AB$ and $CD$ are parallel. Angle $ACD = 15^\circ$, so Angle $BAC = 15^\circ$.
* Angle $D$: Look at triangle $ADC$. The angles must add to $180^\circ$. $180 - 24 - 15 = 141^\circ$.
* Angle $B$: Opposite Angle $D$, so Angle $B = 141^\circ$.
12. Rectangle EFGH
* *Note: The markings show this is a rectangle (diagonals are equal and bisect each other).*
* Side $EH$: Opposite side $FG = 5$, so $EH = 5$.
* Angle $GEH$: Corner angles of a rectangle are $90^\circ$. We know part of it ($\angle FEH$) is not given directly, but we can find parts using triangles. Let's look at the triangle formed by the diagonals.
* Actually, simpler approach: $\angle FEH = 90^\circ$. We are given $\angle FEG = 70^\circ$? No, the arc is for $\angle GEF$? Let's assume the label $70^\circ$ is $\angle GEF$ and $20^\circ$ is $\angle GEH$? Wait, looking closely at vertex E: The angle labeled $70^\circ$ is $\angle FEG$ (part of the corner). The angle labeled $20^\circ$ is at vertex H, specifically $\angle GHE$? No, it looks like $\angle EHG$ is split. Let's look at the diagonal intersection.
* Let's use alternate interior angles. Line $FG$ is parallel to $EH$. Diagonal $EG$ is a transversal. Therefore, $\angle FGE = \angle GEH$.
* Let's look at Vertex E. The corner is $90^\circ$. If the angle labeled $70^\circ$ is $\angle FEG$, then $\angle GEH = 90 - 70 = 20^\circ$.
* Let's check Vertex H. The corner is $90^\circ$. If the angle labeled $20^\circ$ is $\angle EHG$? No, usually these labels refer to the angle between the diagonal and the side. Let's assume the $20^\circ$ is $\angle GHE$? No, that would make triangle $EGH$ have angles $20, 90, 70$. That works.
* So, $\angle GEH = 20^\circ$ (complement of 70).
* $\angle FEG = 70^\circ$.
* $\angle EFG = 90^\circ$ and $\angle FGH = 90^\circ$ and $\angle GHE = 90^\circ$ and $\angle HEF = 90^\circ$.
* Using alternate interior angles: $\angle EGF = \angle GEH = 20^\circ$.
* $\angle FGE = 20^\circ$. Then in triangle $EFG$: $90 + 70 + 20 = 180$. Correct.
* $\angle HEG = 20^\circ$.
* $\angle EGH = 70^\circ$ (alternate interior to $\angle FEG$).
* $\angle FHG$? Diagonal $FH$. Triangle $FGH$ is congruent to $HEF$.
* Let's just list the obvious ones based on the diagram labels which seem to indicate:
* $\angle FEG = 70^\circ$
* $\angle GEH = 20^\circ$ (since $90-70=20$)
* $\angle EFG = 90^\circ, \angle FGH = 90^\circ, \angle GHE = 90^\circ, \angle HEF = 90^\circ$.
* By alternate interior angles: $\angle EGF = 20^\circ$ and $\angle FGE$? No, $\angle EGF$ is the angle at G inside triangle EFG. $\angle EGF = 20^\circ$.
* $\angle HGE = 70^\circ$.
* At the center intersection (let's call it P): Triangle $EPG$ has angles $20, 20, 140$. So vertical angle $\angle FPH = 140^\circ$. Adjacent angle $\angle EPF = 40^\circ$.
13. Parallelogram ABCD with Diagonals
* Given: $DX = 4$ and $AX = 6$.
* Rule: Diagonals bisect each other. This means $X$ is the midpoint of both $BD$ and $AC$.
* Find $BX$: Since $X$ is the midpoint of $BD$, $BX = DX$. So, $BX = 4$.
* Find $BD$: $BD = BX + DX = 4 + 4 = 8$.
* Find $XC$: Since $X$ is the midpoint of $AC$, $XC = AX$. So, $XC = 6$.
* Find $AC$: $AC = AX + XC = 6 + 6 = 12$.
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Final Answer:
1. parallelogram
2. congruent
3. congruent
4. supplementary
5. bisect
6. $\angle M \cong \angle T$ and $\angle A \cong \angle H$
7. $\angle M$ & $\angle A$, $\angle A$ & $\angle T$, $\angle T$ & $\angle H$, $\angle H$ & $\angle M$
8. $MA \cong HT$ and $MT \cong AH$
9. $PQ = 4$, $OP = 6$, $\angle O = 148^\circ$, $\angle P = 32^\circ$, $\angle Q = 148^\circ$
10. $YZ = 7$, $WZ = 10$, $\angle Y = 138^\circ$, $\angle X = 42^\circ$, $\angle Z = 42^\circ$
11. $BC = 13$, $AB = 12$, $\angle ACB = 24^\circ$, $\angle BAC = 15^\circ$, $\angle D = 141^\circ$, $\angle B = 141^\circ$
12. $EH = 5$, All corner angles are $90^\circ$. $\angle GEH = 20^\circ$, $\angle EGF = 20^\circ$, $\angle HGE = 70^\circ$.
13. $BX = 4$, $BD = 8$, $XC = 6$, $AC = 12$
Parent Tip: Review the logic above to help your child master the concept of practice 6 2 properties of parallelograms worksheet answers.
