This geometry parallelograms worksheet answer key solves all 24 problems of the parallelograms and special quadrilaterals worksheet step by step.
Every solution below names the theorem it uses and shows each calculation. Compare your reasoning, not just your final numbers, and use the check step at the end of each solution to catch sign and arithmetic slips. Proof reasons are listed in full, so you can rewrite them in your own words.
Geometry Parallelograms Worksheet Answer Key: problems 1–5
Problem 1: Opposite sides
Opposite sides of a parallelogram are congruent. So \(CD=AB=12\text{ in}\) and \(DA=BC=7\text{ in}\).
CD = 12 in, DA = 7 in
Problem 2: Opposite and consecutive angles
Opposite angles are congruent, so \(m\angle C=m\angle A=68^{\circ}\). Consecutive angles are supplementary, so \(m\angle B=180^{\circ}-68^{\circ}=112^{\circ}\). Then \(m\angle D=m\angle B=112^{\circ}\). Check: \(68+112+68+112=360\).
\(m\angle B=112^{\circ}\), \(m\angle C=68^{\circ}\), \(m\angle D=112^{\circ}\)
Problem 3: Diagonals bisect each other
The diagonals bisect each other, so \(E\) is the midpoint of each. \(AE=\frac{18}{2}=9\) and \(BE=\frac{26}{2}=13\).
\(AE=9\), \(BE=13\)
Problem 4: Sides with a variable
Opposite sides are congruent: \(3x+5=5x-9\). Subtract \(3x\): \(5=2x-9\). Add 9: \(14=2x\), so \(x=7\). Then \(AB=3(7)+5=26\). Check: \(CD=5(7)-9=26\).
\(x=7\), \(AB=26\text{ in}\)
Problem 5: Opposite angles with a variable
Opposite angles are congruent: \(4x+8=6x-12\). Then \(20=2x\) and \(x=10\). So \(m\angle A=4(10)+8=48^{\circ}\) and \(m\angle C=6(10)-12=48^{\circ}\). Consecutive angles are supplementary: \(m\angle B=180^{\circ}-48^{\circ}=132^{\circ}\).
\(x=10\), \(m\angle B=132^{\circ}\)
Geometry Parallelograms Worksheet Answer Key: problems 6–10
Problem 6: Consecutive angles with a variable
\(\angle B\) and \(\angle C\) share side \(\overline{BC}\), so they are consecutive and supplementary: \((3x+10)+(2x+20)=180\). Then \(5x+30=180\), \(5x=150\) and \(x=30\). So \(m\angle B=3(30)+10=100^{\circ}\) and \(m\angle C=2(30)+20=80^{\circ}\). Check: \(100+80=180\).
\(x=30\), \(m\angle B=100^{\circ}\), \(m\angle C=80^{\circ}\)
Problem 7: Pieces of the diagonals
Each diagonal is bisected, so \(AE=EC\): \(2x+3=4x-7\), \(10=2x\), \(x=5\). Then \(AE=13\) and \(AC=2(13)=26\). Also \(BE=ED\): \(3y+1=y+13\), \(2y=12\), \(y=6\). Then \(BE=19\) and \(BD=2(19)=38\).
\(AC=26\), \(BD=38\)
Problem 8: Parallelogram or not?
- Yes. Both pairs of opposite sides are congruent.
- No. Two pairs of congruent adjacent sides describe a kite, which is not a parallelogram in general.
- Yes. A quadrilateral whose diagonals bisect each other is a parallelogram.
- No. Exactly one pair of parallel sides describes a trapezoid.
- Yes. One pair of sides that is both parallel and congruent is enough.
a) yes, b) no, c) yes, d) no, e) yes
Problem 9: Perimeter with two variables
Opposite sides are congruent. From \(AB=CD\): \(2x+3=3x-2\), so \(x=5\) and \(AB=13\). From \(BC=DA\): \(4y-1=2y+7\), \(2y=8\), so \(y=4\) and \(BC=15\). Perimeter \(=2(13+15)=56\).
56 in
Problem 10: Diagonals of a rectangle
The diagonals of a rectangle are congruent: \(5x-4=3x+14\). Then \(2x=18\) and \(x=9\). So \(AC=5(9)-4=41\). Check: \(BD=3(9)+14=41\).
\(x=9\), \(AC=41\)
Geometry Parallelograms Worksheet Answer Key: problems 11–15
Problem 11: Length of a rectangle diagonal
A diagonal and two sides form a right triangle, so \(d^{2}=8^{2}+15^{2}=64+225=289\). Then \(d=\sqrt{289}=17\). The other diagonal is congruent, so it is also 17 cm.
17 cm
Problem 12: Rhombus perimeter and diagonals
The four sides are congruent, so \(AB=\frac{52}{4}=13\). The diagonals bisect each other, so \(AE=12\), and they are perpendicular, so triangle \(AEB\) is a right triangle. Then \(BE^{2}=13^{2}-12^{2}=169-144=25\) and \(BE=5\). So \(BD=2(5)=10\).
\(BD=10\text{ cm}\)
Problem 13: Diagonal bisects an angle of a rhombus
The two halves are equal: \(3x+4=5x-12\), so \(16=2x\) and \(x=8\). Each half is \(3(8)+4=28^{\circ}\), so \(m\angle DAB=2(28)=56^{\circ}\). Consecutive angles are supplementary: \(m\angle ABC=180^{\circ}-56^{\circ}=124^{\circ}\).
\(m\angle DAB=56^{\circ}\), \(m\angle ABC=124^{\circ}\)
Problem 14: Diagonals of a square
A square is a rectangle, so its diagonals are congruent: \(QS=18\). They bisect each other, so \(SE=\frac{18}{2}=9\text{ cm}\). A square is also a rhombus, so the diagonals are perpendicular: \(m\angle PEQ=90^{\circ}\). Diagonal \(\overline{PR}\) bisects the right angle at \(P\), so \(m\angle RPQ=45^{\circ}\).
\(SE=9\text{ cm}\), \(m\angle PEQ=90^{\circ}\), \(m\angle RPQ=45^{\circ}\)
Problem 15: Base angles of an isosceles trapezoid
Base angles are congruent, so \(m\angle B=72^{\circ}\). Angles on the same leg between the parallel bases are supplementary, so \(m\angle D=180^{\circ}-72^{\circ}=108^{\circ}\). The other base angle matches: \(m\angle C=108^{\circ}\). Check: \(72+72+108+108=360\).
\(m\angle B=72^{\circ}\), \(m\angle C=108^{\circ}\), \(m\angle D=108^{\circ}\)
Geometry Parallelograms Worksheet Answer Key: problems 16–20
Problem 16: Diagonals of an isosceles trapezoid
The diagonals of an isosceles trapezoid are congruent: \(4x-3=2x+13\). Then \(2x=16\) and \(x=8\). So \(AC=4(8)-3=29\). Check: \(BD=2(8)+13=29\).
\(x=8\), \(AC=29\)
Problem 17: Angles of a kite
The four angles of a quadrilateral add to \(360^{\circ}\). So \(m\angle Y=360^{\circ}-70^{\circ}-118^{\circ}-118^{\circ}=54^{\circ}\).
\(m\angle Y=54^{\circ}\)
Problem 18: Name every quadrilateral
- Square, rectangle, rhombus, parallelogram, quadrilateral.
- Rhombus, parallelogram, quadrilateral. Bisecting diagonals make it a parallelogram, perpendicular diagonals make it a rhombus, and unequal diagonals rule out a square.
- Trapezoid and quadrilateral (an isosceles trapezoid). It is not a parallelogram because it has only one pair of parallel sides.
a) square (also rectangle, rhombus, parallelogram), b) rhombus, c) isosceles trapezoid
Problem 19: Quilt patch
Opposite sides are congruent, so \(CD=9\) and \(DA=6\). Perimeter \(=2(9+6)=30\text{ in}\). Opposite angles: \(m\angle C=65^{\circ}\). Consecutive angles: \(m\angle B=m\angle D=180^{\circ}-65^{\circ}=115^{\circ}\).
30 in of ribbon; \(m\angle B=115^{\circ}\), \(m\angle C=65^{\circ}\), \(m\angle D=115^{\circ}\)
Problem 20: Is the frame a rectangle?
- Both pairs of opposite sides are congruent (24 in and 24 in, 18 in and 18 in), so the frame is a parallelogram.
- A parallelogram with congruent diagonals is a rectangle, so all four corners are right angles.
- \(18^{2}+24^{2}=324+576=900\) and \(30^{2}=900\), so a right angle between the 18-inch and 24-inch sides is consistent with a 30-inch diagonal.
Parallelogram with congruent diagonals, so a rectangle
Geometry Parallelograms Worksheet Answer Key: problems 21–24
Problem 21: Flying kite frame
The horizontal spar is bisected, so each half is 12 in. The lower part of the vertical spar is \(25-9=16\) in. Top sides: \(\sqrt{9^{2}+12^{2}}=\sqrt{225}=15\) in. Bottom sides: \(\sqrt{16^{2}+12^{2}}=\sqrt{400}=20\) in. The kite has two pairs of congruent adjacent sides, so the tape needed is \(2(15+20)=70\) in.
70 in
Problem 22: Trapezoid garden bed
Perimeter: \(20+12+5+5=42\text{ ft}\). Drop perpendicular segments from the ends of the short base. Each leg overhangs by \(\frac{20-12}{2}=4\) ft, so \(h^{2}+4^{2}=5^{2}\), \(h^{2}=9\) and \(h=3\) ft.
Perimeter 42 ft; distance between bases 3 ft
Problem 23: Rhombus floor tile
\(AB=AD=8\) because the sides of a rhombus are congruent, so triangle \(ABD\) is isosceles with vertex angle \(60^{\circ}\). Its base angles are \(\frac{180-60}{2}=60^{\circ}\) each. All three angles are \(60^{\circ}\), so the triangle is equilateral and \(BD=8\text{ in}\).
\(BD=8\text{ in}\); each angle of triangle ABD measures \(60^{\circ}\)
Problem 24: Proof: diagonals bisect each other
| Statements | Reasons |
|---|---|
| 1. \(ABCD\) is a parallelogram | 1. Given |
| 2. \(\overline{AB}\) is parallel to \(\overline{DC}\) | 2. Definition of a parallelogram |
| 3. \(m\angle BAE=m\angle DCE\) and \(m\angle ABE=m\angle CDE\) | 3. Alternate interior angles theorem (transversals \(\overline{AC}\) and \(\overline{BD}\)) |
| 4. \(AB=CD\) | 4. Opposite sides of a parallelogram are congruent |
| 5. Triangle \(ABE\) is congruent to triangle \(CDE\) | 5. ASA (steps 3 and 4) |
| 6. \(AE=CE\) and \(BE=DE\) | 6. Corresponding parts of congruent triangles are congruent (CPCTC) |
Reasons: Given; definition; alternate interior angles; opposite sides congruent; ASA; CPCTC
How to use this geometry parallelograms worksheet answer key
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