Rhombus problems with detailed solutions. Also see our Rhombus Calculator and Solver.
Definition and Properties
A rhombus is a parallelogram with four congruent sides. A square is a special case of a rhombus.
For rhombus \(ABCD\) above:
- All sides are congruent: \(AB = BC = CD = DA = a\)
- Opposite sides are parallel: \(AD \parallel BC\) and \(AB \parallel DC\)
- Diagonals are perpendicular: \(AC \perp BD\)
- Opposite angles are congruent: \(\angle A = \angle C\) and \(\angle B = \angle D\)
- Consecutive angles are supplementary: \(\angle A + \angle B = 180^\circ\), \(\angle B + \angle C = 180^\circ\), \(\angle C + \angle D = 180^\circ\), \(\angle D + \angle A = 180^\circ\)
Area Formulas
- Base × height: \(A = a \times h\)
- Side and angle: \(A = a^2 \sin \theta\) (where \(\theta\) is any interior angle)
- Diagonals: \(A = \dfrac{1}{2} d_1 d_2\)
Problems & Step-by-Step Solutions
Problem 1: Area from Obtuse and Acute Angle Ratio
Problem Statement: The obtuse angle of a rhombus is twice its acute angle. The side length is 10 feet. Find the area.
Solution:
Let \(x\) = acute angle, then \(2x\) = obtuse angle.
Since consecutive angles are supplementary:
\(x + 2x = 180^\circ\)
\(3x = 180^\circ\)
\(x = 60^\circ\)
Using the area formula: \(A = a^2 \sin \theta\)
\(A = (10)^2 \sin 60^\circ = 100 \times \dfrac{\sqrt{3}}{2} \approx 86.6 \, \text{ft}^2\)
Problem 2: Perimeter from Diagonals
Problem Statement: The diagonals of a rhombus are 20 m and 48 m. Find the perimeter.
Solution:
Let diagonals \(d_1 = 20 \, \text{m}\), \(d_2 = 48 \, \text{m}\).
Diagonals bisect each other at right angles, so half-lengths are 10 m and 24 m.
Using the Pythagorean theorem on one right triangle:
\(a^2 = 10^2 + 24^2 = 100 + 576 = 676\)
\(a = \sqrt{676} = 26 \, \text{m}\) (side length)
Perimeter: \(P = 4a = 4 \times 26 = 104 \, \text{m}\)
Problem 3: Area from Perimeter and One Diagonal
Problem Statement: The perimeter of a rhombus is 120 feet and one diagonal is 40 feet. Find the area.
Solution:
Perimeter = 120 ft ⇒ side length \(a = 120 / 4 = 30 \, \text{ft}\).
One diagonal = 40 ft ⇒ half diagonal = 20 ft.
Using the Pythagorean theorem:
\(30^2 = 20^2 + x^2\) where \(x\) = half of the other diagonal
\(900 = 400 + x^2\)
\(x^2 = 500\)
\(x = \sqrt{500} = 10\sqrt{5} \, \text{ft}\)
Full diagonal: \(d_2 = 2x = 20\sqrt{5} \, \text{ft}\)
Area using diagonals: \(A = \dfrac{1}{2} d_1 d_2 = \dfrac{1}{2} \times 40 \times 20\sqrt{5} = 400\sqrt{5} \, \text{ft}^2\)