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**

- In a rhombus all sides are equal
- In a rhombus opposite angles are equal.
- Also, in a rhombus the sum of adjacent angles are supplementary i.e. (∠B + ∠C = 180°).
- In a rhombus, if one angle is right, then all angles are right.
- In a rhombus, each diagonal of a rhombus divides it into two congruent triangles.
- 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,

d_{₁}, d_{₂} |
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 d_{1} = 30 cm and d_{2} = 16 cm

AO= \(\frac{30}{2}=15 cm\),

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

∠AOB=90^{∘}

From Pythagorean Theorem, we know

AB^{2}=AO^{2}+BO^{2}

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

## Leave a Reply