# Special Parallelograms

A parallelogram is a plane figure with two pairs of opposite sides. The opposite sides are parallel and equal, and the opposite angles are of equal measure. Parallelograms can be equilateral, equiangular, or both. There are three special types of parallelogramsâ€” rectangle, rhombus, and square. They are special because, in addition to the general properties of a parallelogram that they show, they have their unique properties. The unique properties are as follows:

• A rectangle has four right angles. So it is equiangular (with all angles equal)
• A rhombus has four congruent sides. So it is equilateral (all sides equal)
• A square has four right angles and four congruent sides. So a square is equilateral and equiangular.

Now let us take a look at each of their properties closely.

## Properties

Let us consider each of the properties of special parallelograms in the following segments.

### Rectangle

A rectangle is a plane figure with four straight sides making four right internal angles.

1. Has four sides and four angles; in â–­ ABCD, AB, BC, CD, and DA are four sides and âˆ ABC, âˆ BCD, âˆ CDA, âˆ DAB are four angles
2. Opposite sides are equal; so AB = CD and BC= DA
3. Opposite sides are parallel; AB âˆ¥ CD and BC âˆ¥ DA
4. All the angles are 90Â°; in â–­ ABCD, âˆ ABC = âˆ BCD = âˆ CDA =âˆ DAB =  90Â°
5. The diagonals are equal and bisect each other; so AC = BD
6. The sum of the interior angles is equal to 360 degrees; âˆ ABC + âˆ BCD + âˆ CDA  + âˆ DAB = 360Â°

### Rhombus

1. All four sides are equal; in rhombus ABCD, AB = BC = CD = DA
2. Opposite sides are parallel; so AB âˆ¥ CD and BC âˆ¥ DA
3. Opposite angles are equal; âˆ DAB = âˆ BCD and âˆ ABC = âˆ CDA
4. Two diagonals are equal and perpendicular bisectors; so AC = BD & AC âŠ¥ BD
5. Adjacent angles add up 180Â° (supplementary); so âˆ DAB +âˆ ABC = 180Â°, âˆ ABC + âˆ BCD = 180Â°, âˆ BCD + âˆ CDA = 180Â°, and âˆ CDA + âˆ DAB = 180Â°

### Square

1. All four interior angles are right angles; in â–¡ ABCD, âˆ ABC = âˆ BCD = âˆ CDA =âˆ DAB =  90Â°
2. Has four vertices and four sides; A, B, C and D are vertices and AB, BC, CD and DA are sides
3. All four sides are congruent; so AB â‰… BC â‰… CD â‰…DA
4. Opposite sides are parallel to each other; AB âˆ¥ CD and BC âˆ¥ DA
5. Diagonals are equal and perpendicular bisectors; so AC = BD & AC âŠ¥ BD

## Formulas

Let us solve some examples to understand the concept better

## Solved Examples

In rectangle ABCD below, diagonals AC and BD intersect at point R. If AR = 2x – 4 and CR = x + 12, find BD.

Solution:

As we know,
In rectangle, diagonals bisect each other
AR = CR, here AR = 2x – 4 and CR = x + 12
2x – 4 = x + 12 (substituting values of AR & CR)
x = 16
Using either of the given equations to determine BD since each half of the diagonals are equal
BD = 26 units

Given ABCD is a rhombus and âˆ ADC is 60Â°. Find the measure of âˆ CAD.

Solution:

As we know,
In a rhombus, opposite angles are equal,
âˆ´ âˆ ABC = 60Â°
Consecutive angles are supplementary
âˆ´ âˆ BAD = (180 – 60)Â°
= 120Â°
Diagonals are angle bisector
= 60Â°

For square ABCD, state whether the following statements are true or false.
1. The length of AC is equal to the length of BD.
2. The diagonals AC and BD bisect each other at right angles.
3. The perimeter of the above square could be given as 4CD.

Solution:

As we know,
Diagonals of a square are congruent,
âˆ´ Statement 1: The length of AC is equal to the length of BD is True
As we know,
Diagonals of a square always bisect each other at right angles
âˆ´ Statement 2: The diagonals AC and BD bisect each other at right angles is True
As we know,
Perimeter of square is 4a, if any side is ‘a’ units
âˆ´ Statement 3: The perimeter of the above square could be given as 4CD is True