Rhombus Formula

A rhombus is a 2-dimensional shape that has four equal sides. Rhombus consists of all sides equal and its opposite angles are equal in measure. Let us now discuss the rhombus formula i.e. area and perimeter of the rhombus.

Rhombus Formula

What is a Rhombus?

Rhombus is a special type of parallelogram that has all sides equal. Rhombus is quadrilateral whose all sides are equal.

Properties of Rhombus

1. In a rhombus all sides are equal
2. In a rhombus opposite angles are equal.
3. Also, in a rhombus the sum of adjacent angles are supplementary i.e. (∠B + ∠C = 180°).
4. In a rhombus, if one angle is right, then all angles are right.
5. In a rhombus, each diagonal of a rhombus divides it into two congruent triangles.
6. Diagonals of a rhombus bisect each other and also perpendicular to each other.

The Perimeter of a Rhombus

The perimeter is the sum of the length of all the 4 sides. In rhombus all sides are equal.

So, Perimeter of rhombus = 4 × side 

P = 4s

Where,

s	length of a side of a rhombus

Area of Rhombus

The area of a rhombus is the number of square units inside the polygon. The area of a rhombus can be determined in two ways:

i) By multiplying the base and height as rhombus is a special type of parallelogram.

Area of rhombus = b  × h

Where,

b	Base of Rhombus
h	Height of the Rhombus

ii) By finding the product of the diagonal of the rhombus and divide the product by 2.

Area of rhombus = \(\frac{1}{2}\)  × d_₁  × d_₂

where,

d₁, d₂

Diagonals of Rhombus

Derivation of Area of Rhombus

Let ABCD is a rhombus whose base AB = b, DB ⊥ AC,   DB is diagonal of rhombus = d_₁, AC is diagonal of rhombus = d_₂, and the altitude from C on AB is CE, i.e., h.

i) Area of rhombus ABCD   = 2 Area of ∆ ABC

= 2  ×  \(\frac{1}{2}\) AB  × CD sq units.

= 2  ×  \(\frac{1}{2}\) b  × h sq. units

= base × height sq. units

ii) Area of rhombus    = 4  × area of ∆ AOB

= 4  ×  \(\frac{1}{2}\)  ×  AO  ×  OB sq. units

= 4  ×  \(\frac{1}{2}\)  × \(\frac{1}{2}\) d_₂  ×  \(\frac{1}{2}\) d_₁ sq. units

so,

= 4  ×  \(\frac{1}{8}\) d_₁  ×  d_₂ square units

= \(\frac{1}{2}\)  ×  d_₁  × d_₂

Therefore, area of rhombus = \(\frac{1}{2} \)(product of diagonals) square units

Solved Examples

Q.1. What is the perimeter of a rhombus ABCD whose diagonals are 16 cm and 30 cm ?

Solution: Given d₁ = 30 cm and d₂ = 16 cm
AO= \(\frac{30}{2}=15 cm\),

BO= \(\frac{16}{2}=8 cm\),

∠AOB=90^∘

From Pythagorean Theorem, we know
AB²=AO²+BO²
AB =\(\sqrt{289}\)

=17 cm

Since, AB=BC=CD=DA,
Perimeter of ABCD = 17 × 4 = 68 cm

Q.2. Find the area of the rhombus having each side equal to 17 cm and one of its diagonals equal to 16 cm.

Solution: In rhombus ABCD, AB = BC = CD = DA = 17 cm
AC = 16 cm, AO = 8 cm

In ∆ AOD,

AD² = AO² + OD²

17² = 8² + OD²

289 = 64 + OD²

225 = OD²

OD = 15 cm

Therefore, BD   = 2 OD

= 2  × 15

= 30 cm

Now, area of rhombus = \(\frac{1}{2}\)  ×  d_₁  × d_₂

= \(\frac{1}{2} \)× 16  × 30

= 240 cm²