6.3 Special parallelograms: rectangles, rhombuses, and squares

Properties and Theorems of Special Parallelograms

Every special parallelogram is, first and foremost, a parallelogram. That means it inherits all the standard parallelogram properties (opposite sides parallel and congruent, opposite angles congruent, consecutive angles supplementary, diagonals bisect each other). Rectangles, rhombuses, and squares each add extra constraints on top of those, and understanding exactly which constraints stack on top of which is the key to this section.

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Properties of Special Quadrilaterals

Rectangles are parallelograms where every angle is a right angle (90°).

• Opposite sides are parallel and congruent
• All four angles are 90°
• Diagonals are congruent and bisect each other

Because the diagonals are congruent and bisect each other, each half-diagonal has the same length. This creates four congruent segments radiating from the center, which means the four triangles formed by the diagonals are isosceles.

Rhombuses are parallelograms where all four sides are congruent.

• All sides are congruent
• Opposite sides are parallel
• Opposite angles are congruent (but not necessarily 90°)
• Diagonals are perpendicular and bisect each other
• Each diagonal bisects a pair of opposite angles

That last property is easy to overlook. If a rhombus has a 60° vertex angle, each diagonal through that vertex splits it into two 30° angles.

Squares satisfy the definitions of both a rectangle and a rhombus, so they inherit every property from both lists.

• All sides congruent
• All angles 90°
• Diagonals are congruent, perpendicular, and bisect each other
• Each diagonal bisects a pair of opposite angles (each corner splits into two 45° angles)

Theorems for Special Parallelograms

Theorem: The diagonals of a rectangle are congruent.

You can prove this with SSS or SAS. Consider rectangle $ABCD$ with diagonals $\overline{AC}$ and $\overline{BD}$. Look at triangles $\triangle ABC$ and $\triangle DCB$:

1. $\overline{AB} \cong \overline{DC}$ (opposite sides of a rectangle)
2. $\overline{BC} \cong \overline{BC}$ (shared side)
3. $\angle ABC \cong \angle DCB$ (both are 90°)
4. By SAS, $\triangle ABC \cong \triangle DCB$
5. Therefore $\overline{AC} \cong \overline{BD}$ (CPCTC)

Theorem: The diagonals of a rhombus are perpendicular.

Consider rhombus $ABCD$ with diagonals intersecting at point $E$. Since the diagonals of any parallelogram bisect each other, $E$ is the midpoint of both diagonals.

1. In $\triangle AEB$, we know $\overline{AE} \cong \overline{AE}$ (reflexive), $\overline{AB} \cong \overline{AD}$ (all sides of a rhombus are congruent), and $\overline{BE} \cong \overline{DE}$ (diagonals bisect each other).
2. By SSS, $\triangle AEB \cong \triangle AED$.
3. Therefore $\angle AEB \cong \angle AED$ (CPCTC).
4. Since $\angle AEB$ and $\angle AED$ are supplementary (they form a linear pair) and congruent, each must be 90°.
5. The diagonals are perpendicular.

Theorem: Each diagonal of a rhombus bisects a pair of opposite angles.

This follows directly from the triangle congruences above. Since $\triangle AEB \cong \triangle AED$, we get $\angle BAE \cong \angle DAE$, so diagonal $\overline{AC}$ bisects $\angle A$. The same reasoning applies at every vertex.

Problem-Solving with Special Quadrilaterals

The diagonals of a rhombus are perpendicular and bisect each other, which means they cut the rhombus into four right triangles. This is the setup you'll use most often.

Example: A rhombus has a side length of 5 units and one diagonal of length 8 units. Find the other diagonal.

1. The diagonals bisect each other, so the 8-unit diagonal splits into two segments of 4 units each.

2. Each of the four right triangles has a hypotenuse of 5 (the side of the rhombus) and one leg of 4 (half the known diagonal).

3. Apply the Pythagorean theorem to find the other leg:

   • $4^2 + b^2 = 5^2$
   • $16 + b^2 = 25$
   • $b^2 = 9$
   • $b = 3$

4. That leg is half the other diagonal, so the full diagonal is $2 \times 3 = 6$ units.

Construction of Special Parallelograms

Constructing a rectangle given two adjacent side lengths:

1. Draw a segment $\overline{AB}$ with the length of one side.
2. Construct a perpendicular line at point $A$.
3. Construct a perpendicular line at point $B$.
4. Mark the length of the second side along each perpendicular (going the same direction) to get points $D$ and $C$.
5. Connect $D$ to $C$. Quadrilateral $ABCD$ is your rectangle.

Constructing a rhombus given the lengths of both diagonals:

1. Draw one diagonal and mark its midpoint.
2. Through the midpoint, construct a perpendicular line.
3. Along that perpendicular, mark half the second diagonal's length on each side of the midpoint.
4. Connect the four endpoints of the diagonals in order to form the rhombus.

Constructing a square given one side length:

1. Draw a segment $\overline{AB}$ with the given side length.
2. Construct perpendicular lines at $A$ and at $B$.
3. Mark the same side length along each perpendicular (same direction) to get points $D$ and $C$.
4. Connect $D$ to $C$. Quadrilateral $ABCD$ is your square.

Applying Special Parallelogram Properties

Finding Measurements from Diagonals

Example: A square has a diagonal of length 10 units. Find the side length.

The diagonals of a square are perpendicular and bisect each other, so they create four right triangles. But there's a faster approach here: each diagonal of a square relates to the side by the formula $d = s\sqrt{2}$.

• $10 = s\sqrt{2}$
• $s = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07$ units

You can verify this with the Pythagorean theorem. The two legs of the right triangle are both $s$, and the hypotenuse is the full diagonal (10):

• $s^2 + s^2 = 10^2$
• $2s^2 = 100$
• $s^2 = 50$
• $s = \sqrt{50} = 5\sqrt{2}$

Real-World Application

Example: A rectangular garden has a perimeter of 60 meters, and its length is 5 meters longer than its width. Find the dimensions.

1. Let the width be $w$ meters, so the length is $w + 5$ meters.
2. Use the perimeter formula: $P = 2l + 2w$
3. $60 = 2(w + 5) + 2w$
4. $60 = 2w + 10 + 2w$
5. $60 = 4w + 10$
6. $50 = 4w$
7. $w = 12.5$

The width is 12.5 meters and the length is 17.5 meters. Quick check: $2(12.5) + 2(17.5) = 25 + 35 = 60$. ✓