Properties of Rhombuses, Rectangles, and Squares

The three special parallelograms rhombus, rectangle, and square — are so-called because they’re special cases of the parallelogram. (In addition, the square is a special case or type of both the rectangle and the rhombus.)

The three-level hierarchy you see with

in the above quadrilateral family tree works just like

A dog is a special type of a mammal, and a Dalmatian is a special type of a dog.

Here are the properties of the rhombus, rectangle, and square. Note that because these three quadrilaterals are all parallelograms, their properties include the parallelogram properties.

• The rhombus has the following properties:

  1. All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite angles are congruent, and consecutive angles are supplementary).
  2. All sides are congruent by definition.
  3. The diagonals bisect the angles.
  4. The diagonals are perpendicular bisectors of each other.

• The rectangle has the following properties:

  • All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other).
  • All angles are right angles by definition.
  • The diagonals are congruent.
• The square has the following properties:

  • All the properties of a rhombus apply (the ones that matter here are parallel sides, diagonals are perpendicular bisectors of each other, and diagonals bisect the angles).
  • All the properties of a rectangle apply (the only one that matters here is diagonals are congruent).
  • All sides are congruent by definition.
  • All angles are right angles by definition.

Now try working through a problem. Given the rectangle as shown, find the measures of angle 1 and angle 2:

Here’s the solution: MNPQ is a rectangle, so angle Q = 90°. Thus, because there are 180° in a triangle, you can say

Now plug in 14 for all the x ’s.

Now find the perimeter of rhombus RHOM .

Here’s the solution: All the sides of a rhombus are congruent, so HO equals x + 2. And because the diagonals of a rhombus are perpendicular, triangle HBO is a right triangle. You finish with the Pythagorean Theorem:

Combine like terms and set equal to zero:

Factor:

( x – 3)( x + 1) = 0

Use Zero Product Property:

x – 3 = 0 or x + 1 = 0

x = 3 or x = –1

You can reject x = –1 because that would result in triangle HBO having legs with lengths of –1 and 0.