Rhombus Problems

Properties of a Rhombus

These are some of the most important properties of a rhombus.

Consider the rhombus ABCD shown in the figure above.

1 - All sides are congruent (equal lengths).

length AB = length BC = length CD = length DA = a.

2 - Opposite sides are parallel.

AD is parallel to BC and AB is parallel to DC.

3 - The two diagonals are perpendicular.

AC is perpendicular to BD.

4 - Opposite internal angles are congruent (equal sizes).

internal angle A = internal angle C
and
internal angle B = internal angle D.

5 - Any two consecutive internal angles are supplementary : they add up to 180 degrees.

angle A + angle B = 180 degrees
angle B + angle C = 180 degrees
angle C + angle D = 180 degrees
angle D + angle A = 180 degrees

Area of a Rhombus

These are three formulas for the area of the rhombus.

formula 1:

area = a*h , where a is the side length of the rhombus and h is the perpendicular distance between two parallel sides of the rhombus.

formula 2:

area = a ²*sin (A) = a ²*sin (B). Since angles A and B are supplementary angles, sin (A) = sin (B).

formula 3:

area = (1/2)*d1*d2, where d1 and d2 are the lengths of the two diagonals.

We now present some problems with detailed solutions.

Problem 1: The size of the obtuse angle of a rhombus is twice the size of its acute angle. The side length of the rhombus is equal to 10 feet. Find its area.

Solution to Problem 1:

• A rhombus has 2 congruent opposite acute angles and two congruent opposite obtuse angles. One of the properties of a rhombus is that any two internal consecutive angles are supplementary. Let x be the acute angle. The obtuse angle is twice: 2x. Which gives the following equation.
  
  x + 2 x = 180 degrees.
  
  
• Solve the above equation for x.
  
  3x = 180 degrees.
  
  x = 60 degrees.
  
  
• We use the formula for the area of a triangle that uses the side lengths and any one of the angles then multiply the area by 2.
  
  area of rhombus = 2 (1 / 2) (10 feet) ² sin (60 degrees)
  
  = 86.6 feet ² (rounded to 1 decimal place)

Problem 2: The lengths of the diagonals of a rhombus are 20 and 48 meters. Find the perimeter of the rhombus?

Solution to Problem 2:

• Below is shown a rhombus with the given diagonals. Consider the right triangle BOC and apply Pythagora's theorem as follows
  
  
  
  BC ² = 10 ² + 24 ²
  
  
• and evaluate BC
  
  BC = 26 meters.
  
  
• We now evaluate the perimeter P as follows:
  
  P = 4 * 26 = 104 meters.
  
  

Problem 3: The perimeter of a rhombus is 120 feet and one of its diagonal has a length of 40 feet. Find the area of the rhombus.

Solution to Problem 3:

• A perimeter of 120 when divided by 4 gives the side of the rhombus 30 feet. The length of the side OC of the right traingle is equal to half the diagonal: 20 feet. Let us now consider the right triangle BOC and apply Pythagora's theorem to find the length of side BO.
  
  
  
  30 ² = BO ² + 20 ²
  
  BO = 10 sqrt(5) feet
  
  
• We now calculate the area of the right triangle BOC and multiply it by 4 to obtain the area of the rhombus.
  
  area = 4 ( 1/2) BO * OC = 4 (1/2)10 sqrt (5) * 20
  
  = 400 sqrt(5) feet ²

More references on geometry.