Parallelograms and Rectangles
A parallelogram has two pairs of parallel sides. From that ONE rule, a whole list of properties follow — equal opposite sides, equal opposite angles, diagonals that bisect each other. Add four right angles and you get a rectangle, which also has equal diagonals.
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Draw a parallelogram on grid paper (slanted, NOT a rectangle). Measure its two pairs of opposite sides. What do you notice? Now measure both diagonals and the angles at each end. Write down everything that looks "equal" or "symmetric".
A parallelogram is a quadrilateral with two pairs of parallel sides. That's the defining property. From it, four other properties automatically follow: opposite sides are equal in length, opposite angles are equal, co-interior angles are supplementary (add to $180^{\circ}$), and the diagonals bisect each other (cut each other in half).
A parallelogram $ABCD$ has $AB \parallel DC$ and $AD \parallel BC$. From this we get: $AB = DC$ and $AD = BC$ (opposite sides equal), $\angle A = \angle C$ and $\angle B = \angle D$ (opposite angles equal), and the diagonals $AC$ and $BD$ meet at a point $M$ where $AM = MC$ and $BM = MD$.
Know
- The 5 properties of a parallelogram
- The extra 2 properties of a rectangle
- How rectangles fit inside the parallelogram family
- Standard angle-property names ("opp. angles of parallelogram", "co-int. angles")
Understand
- Why opposite sides of a parallelogram must be equal
- Why the diagonals always bisect each other
- Why a rectangle is a parallelogram but a parallelogram isn't always a rectangle
Can Do
- Find missing side lengths in a parallelogram using opposite-sides equal
- Find missing angles using opposite-angles or co-interior
- Solve diagonal problems using the bisection property
Wrong: "The diagonals of a parallelogram are equal." NO — only in a rectangle. In a general parallelogram, the diagonals usually have DIFFERENT lengths.
Right: The diagonals of a parallelogram bisect each other, but they aren't equal in length unless the parallelogram is a rectangle.
Wrong: "All four angles of a parallelogram are equal." NO — only the OPPOSITE angles. Adjacent angles are usually different (they sum to $180^{\circ}$).
Right: $\angle A = \angle C$ and $\angle B = \angle D$. If $\angle A = 70^{\circ}$, then $\angle B = 110^{\circ}$ (co-interior).
Book notes · Card 4
- Diagonals of a parallelogram BISECT each other — they do NOT have to be equal.
- Only OPPOSITE angles of a parallelogram are equal; adjacent angles are co-interior (sum $180^{\circ}$).
- If diagonals ARE equal ⇒ the parallelogram is a rectangle.
True or False: In every parallelogram, the two diagonals have equal lengths.
A rectangle is a parallelogram with one extra rule: all four angles are right angles. Because a rectangle is a parallelogram, it inherits ALL five parallelogram properties. Plus it gets two extras: (i) all angles equal $90^{\circ}$, and (ii) the diagonals are EQUAL in length (and still bisect each other).
In rectangle $PQRS$: opposite sides equal & parallel (parallelogram property), $\angle P = \angle Q = \angle R = \angle S = 90^{\circ}$, and $PR = QS$ (diagonals equal length AND bisect each other at midpoint $M$).
Book notes · Card 5
- Rectangle = parallelogram + 4 right angles ⇒ gets equal diagonals as a bonus.
- All 5 parallelogram properties + 2 rectangle extras (all $90^{\circ}$ angles, equal diagonals).
- Midpoint $M$ of the diagonals satisfies $PM = QM = RM = SM$.
Which extra property does a rectangle have that a non-rectangular parallelogram does NOT have?
When you see a parallelogram or rectangle in a question, scan the diagram and ask: which property gives me the unknown? Match the situation to the right rule and ALWAYS write a reason in brackets — e.g. (opp. sides of parallelogram), (opp. angles of parallelogram), (co-int. angles, $AB \parallel DC$), (diagonals of rectangle equal).
Decision flow for parallelograms:
• Need an opposite side → opp. sides equal
• Need an opposite angle → opp. angles equal
• Need an adjacent angle → co-int. angles sum $180^{\circ}$
• Need a half-diagonal → diagonals bisect
• Rectangle only: full diagonal → diagonals equal
Book notes · Card 6
- Match the unknown to the right property: side → opp. sides; angle → opp. or co-int.
- Co-interior angles in a parallelogram always sum to $180^{\circ}$.
- Always state a reason like "(opp. angles of parallelogram)" in brackets.
In parallelogram $ABCD$, if $\angle A = 65^{\circ}$, then $\angle B = $ $^{\circ}$ because adjacent angles are co-interior.
Watch Me Solve It · 3 examples
- 1Opposite angles are equal$\angle C = \angle A = 75^{\circ}$ (opp. angles of parallelogram)
- 2Adjacent angles are co-interior$\angle B = 180 - 75 = 105^{\circ}$ (co-int. angles, $AB \parallel DC$)
- 3Final angle$\angle D = \angle B = 105^{\circ}$ (opp. angles of parallelogram)Check: $75 + 105 + 75 + 105 = 360^{\circ}$ ✓
- 1Use bisection of $PR$$M$ is the midpoint of $PR$ (diagonals of parallelogram bisect), so $MR = PM = 9$.
- 2Find $PR$$PR = PM + MR = 9 + 9 = 18$ cm.
- 3Find $QS$$M$ is the midpoint of $QS$ too, so $MS = QM = 6$, giving $QS = 12$ cm.Note: $PR \ne QS$ in general — only EQUAL if it's a rectangle.
- 1Diagonals of rectangle equal$XZ = WY = 26$ cm (diagonals of rectangle equal).
- 2Diagonals bisect$M$ is the midpoint of $XZ$, so $MX = \tfrac{1}{2} \times XZ$.
- 3Compute$MX = \tfrac{26}{2} = 13$ cm.All four "half diagonals" $WM, MY, XM, MZ$ equal $13$ cm in a rectangle.
Common Pitfalls
Parallelogram (5)
- 2 pairs parallel sides
- Opposite sides equal
- Opposite angles equal
- Co-int. angles $= 180^{\circ}$
- Diagonals bisect each other
Rectangle extras (2)
- All 4 angles $= 90^{\circ}$
- Diagonals are equal in length
- (still bisect each other)
Reasons
- (opp. sides of parallelogram)
- (opp. angles of parallelogram)
- (co-int. angles, $AB \parallel DC$)
- (diagonals of rectangle equal)
Family
- Rectangle is a parallelogram
- Parallelogram is NOT always a rectangle
- If diagonals equal ⇒ rectangle
How are you completing this lesson?
Brain Trainer · 4 problems
Four quick drills using the parallelogram and rectangle properties. Solve, then reveal.
-
1 In parallelogram $ABCD$, $\angle A = 110^{\circ}$. Find $\angle B$.
Co-int. with $\angle A$: $180 - 110$.$\angle B = 70^{\circ}$ -
2 Parallelogram $PQRS$ has $PQ = 12$ cm. Find $SR$.
Opp. sides equal.$SR = 12$ cm -
3 Rectangle $WXYZ$ has $WY = 20$ cm. Find $XZ$.
Diagonals of rectangle are equal.$XZ = 20$ cm -
4 Diagonals of parallelogram meet at $M$ with $AM = 5$ cm. Find $MC$.
Diagonals bisect each other.$MC = 5$ cm
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Parallelogram $EFGH$ has $\angle E = 55^{\circ}$ and $EF = 9$ cm.
(a) Find $\angle F$, giving a reason.
(b) Find $\angle G$, giving a reason.
(c) Find $HG$, giving a reason.
Q7. The diagonals of rectangle $JKLM$ meet at $N$. $JL = 24$ cm.
(a) Find $KM$, giving a reason.
(b) Find $KN$, giving a reason.
(c) Explain why all four "half-diagonals" $JN, KN, LN, MN$ are equal.
Q8. In parallelogram $ABCD$, $\angle A = (3x + 10)^{\circ}$ and $\angle B = (5x - 30)^{\circ}$.
(a) Set up an equation using a parallelogram property.
(b) Solve for $x$.
(c) State all four angles.
Quick Check
1. C — Opposite sides are equal in length.
2. B — $180 - 68 = 112^{\circ}$ (co-interior).
3. D — Diagonals are equal in length.
4. A — $MC = 7$ cm (diagonals bisect each other).
5. B — $\angle C = 105^{\circ}$ (opp. angles of parallelogram).
Show Your Working Model Answers
Q6 (3 marks): (a) $\angle F = 180 - 55 = 125^{\circ}$ (co-int. angles, $EF \parallel HG$) [1]. (b) $\angle G = \angle E = 55^{\circ}$ (opp. angles of parallelogram) [1]. (c) $HG = EF = 9$ cm (opp. sides of parallelogram) [1].
Q7 (3 marks): (a) $KM = JL = 24$ cm (diagonals of rectangle equal) [1]. (b) $KN = \tfrac{KM}{2} = 12$ cm (diagonals bisect each other) [1]. (c) Rectangle has equal diagonals AND they bisect, so all four halves measure $12$ cm [1].
Q8 (3 marks): (a) $\angle A + \angle B = 180^{\circ}$ (co-int. angles): $(3x + 10) + (5x - 30) = 180$ [1]. (b) $8x - 20 = 180 \Rightarrow 8x = 200 \Rightarrow x = 25$ [1]. (c) $\angle A = 85^{\circ}, \angle B = 95^{\circ}, \angle C = 85^{\circ}, \angle D = 95^{\circ}$ [1].
Prove It
In parallelogram $ABCD$, the diagonals meet at $M$. (a) Using alternate angles and congruent triangles, explain why $AM = MC$. (b) A particular parallelogram has $AC = 14$ cm and $BD = 18$ cm. Could it be a rectangle? Why or why not? (c) If you're told $AC = BD$, what kind of parallelogram must it be?
Reveal solution
(a) Triangles $AMB$ and $CMD$: $AB = CD$ (opp. sides), $\angle BAM = \angle DCM$ (alt. angles, $AB \parallel DC$), $\angle ABM = \angle CDM$ (alt. angles). So $\triangle AMB \equiv \triangle CMD$ (AAS), giving $AM = CM$. (b) NO — a rectangle has EQUAL diagonals, but $14 \ne 18$. So this is a non-rectangular parallelogram. (c) Equal diagonals in a parallelogram force all four angles to be $90^{\circ}$, so it must be a rectangle.
Parallelogram
2 pairs of parallel sides.
Opp. sides equal
$AB = DC$ and $AD = BC$.
Opp. angles equal
$\angle A = \angle C$, $\angle B = \angle D$.
Co-int. angles
Adjacent angles sum to $180^{\circ}$.
Diagonals bisect
They cut each other in half (but not always equal).
Rectangle extras
All angles $90^{\circ}$ AND equal diagonals.
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