← Back to the Parallelograms Special Quadrilaterals worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-CO.C.11 · Level: Grade 10 Geometry
Problem 1 Opposite sides of a parallelogram
Opposite sides are congruent: CD = AB = 14.5 and AD = BC = 9.
Perimeter = 2(14.5 + 9) = 47.
CD = 14.5, AD = 9, perimeter = 47
Problem 2 Angles of a parallelogram
Opposite angles are congruent, so m∠C = 68°. Consecutive angles are supplementary, so m∠B = 180° − 68° = 112°, and m∠D = 112°.
∠B = 112°, ∠C = 68°, ∠D = 112°
Problem 3 Diagonals that bisect
The diagonals bisect each other, so CE = AE = 6.5 and DE = BE = 9.25.
AC = 2(6.5) = 13 and BD = 2(9.25) = 18.5.
CE = 6.5, DE = 9.25, AC = 13, BD = 18.5
Problem 4 Opposite sides with algebra
Opposite sides are congruent: 4x − 7 = 2x + 9, so 2x = 16 and x = 8.
AB = 4(8) − 7 = 25, and CD = 2(8) + 9 = 25.
x = 8; AB = 25
Problem 5 Consecutive angles with algebra
Consecutive angles are supplementary: (4x + 6) + (6x − 16) = 180, so 10x − 10 = 180 and x = 19.
∠A = 4(19) + 6 = 82° and ∠B = 6(19) − 16 = 98°. Check: 82 + 98 = 180.
x = 19; 82° and 98°
Problem 6 Diagonals with algebra
The diagonals bisect each other: 3x + 1 = 5x − 9, so 10 = 2x and x = 5.
AE = 16, so AC = 2(16) = 32.
x = 5; AC = 32
Problem 7 Is it a parallelogram
- Both pairs of opposite sides are congruent: yes.
- Both pairs of opposite angles are congruent: yes.
- No. An isosceles trapezoid has this pattern and is not a parallelogram.
- Yes: diagonals that bisect each other prove a parallelogram.
(a) yes, (b) yes, (c) no, (d) yes
Problem 8 Make it a parallelogram
Both pairs of opposite sides must be congruent. First pair: 2x + 5 = 4x − 9, so 14 = 2x and x = 7. Second pair: 3y − 1 = y + 7, so 2y = 8 and y = 4.
Check: AB = CD = 19 and AD = BC = 11.
x = 7, y = 4
Problem 9 Parallelogram on the plane
Midpoint of AC: ((1 + 8) ÷ 2, (2 + 7) ÷ 2) = (4.5, 4.5). Midpoint of BD: ((6 + 3) ÷ 2, (3 + 6) ÷ 2) = (4.5, 4.5).
The diagonals share a midpoint, so they bisect each other.
Yes, ABCD is a parallelogram
Problem 10 Rectangle diagonals
Rectangle diagonals are congruent: 5x − 3 = 3x + 11, so 2x = 14 and x = 7.
AC = 5(7) − 3 = 32, and BD = 3(7) + 11 = 32.
x = 7; both diagonals are 32
Problem 11 Diagonal of a rhombus
The four sides are equal: 52 ÷ 4 = 13. The diagonals are perpendicular and bisect each other, so a right triangle has legs 5 (half of 10) and half of the other diagonal, with hypotenuse 13.
Half the other diagonal = √(13² − 5²) = √144 = 12, so the diagonal is 24. Area = ½ · 10 · 24 = 120.
Side 13; other diagonal 24; area 120
Problem 12 Angles of a rhombus
Consecutive angles are supplementary: m∠ABC = 180° − 64° = 116°.
Diagonals bisect the angles of a rhombus: m∠CAB = 64° ÷ 2 = 32° and m∠ABD = 116° ÷ 2 = 58°.
Check in the triangle formed by the diagonals: 32° + 58° + 90° = 180°.
116°, 32° and 58°
Problem 13 Trapezoid midsegment
The midsegment is the average of the bases: 3x + 11 = ½[(4x + 2) + (6x − 8)] = 5x − 3. So 14 = 2x and x = 7.
The bases are 4(7) + 2 = 30 and 6(7) − 8 = 34, and the midsegment is 3(7) + 11 = 32, the average of 30 and 34.
x = 7; bases 30 and 34; midsegment 32
Problem 14 Isosceles trapezoid
Base angles on the same base are congruent, so two angles are 72°. Angles between a leg and different bases are supplementary: 180° − 72° = 108°, so the other two angles are 108°.
Diagonals of an isosceles trapezoid are congruent: 2x + 9 = 4x − 7, so 16 = 2x and x = 8. Each diagonal is 25.
Angles 72°, 72°, 108°, 108°; diagonals 25
Problem 15 Building a kite
The crossbar is bisected by the spine, so each half is 12 inches. The top triangle has hypotenuse 15 and leg 12, so the top part of the spine is √(15² − 12²) = 9. The bottom triangle has hypotenuse 20 and leg 12, so the bottom part is √(20² − 12²) = 16.
Spine = 9 + 16 = 25 inches. Area = ½ · 24 · 25 = 300 square inches.
Spine 25 in; area 300 in²
Problem 16 Garden bed fence
The opposite corner is 65°. The two other corners are supplementary to 65°: 180° − 65° = 115°.
Perimeter = 2(18 + 11.5) = 59 feet. Cost = 59 × 8.50 = $501.50.
Angles 65°, 115°, 115°; perimeter 59 ft; cost $501.50
Problem 17 Prove the diagonals bisect each other
| Statement | Reason |
|---|---|
| 1. ABCD is a parallelogram | Given |
| 2. AB ∥ CD and AB ≅ CD | Definition; opposite sides are congruent |
| 3. ∠BAE ≅ ∠DCE and ∠ABE ≅ ∠CDE | Alternate interior angles (parallel lines) |
| 4. △ABE ≅ △CDE | ASA |
| 5. AE ≅ CE and BE ≅ DE | CPCTC |
The diagonals bisect each other
Problem 18 Prove the converse
| Statement | Reason |
|---|---|
| 1. AB ≅ CD and AD ≅ BC | Given |
| 2. AC ≅ AC | Reflexive Property |
| 3. △ABC ≅ △CDA | SSS |
| 4. ∠BAC ≅ ∠DCA and ∠BCA ≅ ∠DAC | CPCTC |
| 5. AB ∥ CD and BC ∥ AD | Converse of the Alternate Interior Angles Theorem |
| 6. ABCD is a parallelogram | Definition of parallelogram |
ABCD is a parallelogram
Problem 19 Prove rectangle diagonals congruent
| Statement | Reason |
|---|---|
| 1. PQRS is a rectangle | Given |
| 2. PQ ≅ SR | A rectangle is a parallelogram, so opposite sides are congruent |
| 3. QR ≅ RQ | Reflexive Property |
| 4. ∠PQR ≅ ∠SRQ | Both are right angles |
| 5. △PQR ≅ △SRQ | SAS |
| 6. PR ≅ SQ | CPCTC |
PR ≅ QS
Problem 20 Coordinate proof of a rhombus
Side lengths: AB = √(4² + 3²) = 5, BC = 9 − 4 = 5, CD = √((9 − 5)² + 3²) = 5, DA = 5 − 0 = 5. All four sides are equal, so the quadrilateral is a rhombus.
Slope of AC = 3 ÷ 9 = 1/3. Slope of BD = (0 − 3) ÷ (5 − 4) = −3. The product (1/3)(−3) = −1, so the diagonals are perpendicular.
Rhombus with all sides 5; diagonals perpendicular (slopes 1/3 and −3)