# What is Rhombus?

A **Rhombus** is a Parallelogram whose four sides all have the same length. Also called an equilateral quadrilateral. The figure represents a Rhombus.

**Properties**

- All sides of the rhombus are equal.
- The opposite sides of a rhombus are parallel.
- The opposite angles of a rhombus are equal.

**Properties of Diagonals of Rhombus**

- In a rhombus, diagonals bisect each other at right angles.
- Diagonals bisect the angles of a rhombus.

**To Prove**

**1. If a quadrilateral is a rhombus, then the diagonals are perpendicular bisectors of each other.**

**First, we will prove that the diagonals bisect each other then we will prove that the diagonals are perpendicular to each other.**

**(a) Diagonals bisect each other**

**Proof:**

Since the rhombus is a parallelogram so the opposite sides are parallel and equal to each other.

So in the triangle AOD and COB we have

AD = CB ( opposite sides)

∠ADO = ∠CBO (opposite angles)

∠DAO = ∠BCO (opposite angles)

Hence both the triangles are congruent.

Hence AO = CO and DO = BO.

Hence the diagonals bisect each other.

**(b) Diagonals are perpendicular to each other**

**Proof:**

Given that ABCD is a rhombus

AB = AD (definition of rhombus)

AO = AO (Common side)

BO = OD (diagonals bisect each other as proved above)

△AOD ≅ △AOB (Side-Side-Side postulate)

∠AOD ≅ ∠AOB (Corresponding angles in congruent triangles )

Also ∠AOD + ∠AOB = 180

Hence AC ⊥ DB.

Hence the diagonals are the perpendicular bisector of each other.

**2. The area of the rhombus** **is equal to half of the product of the diagonals.**

**Proof:**

Given that ABCD is a rhombus the diagonal AC and BD cut at point O.

Now we know that ∠AOD = ∠AOB = ∠COD = ∠BOC = 90 degree.

Hence we can see that the rhombus ABCD is equally into four parts.

So the area of rhombus ABCD = area of triangle AOD + area of triangle AOB + area of triangle BOC + area of triangle COD.1/2 ×

AO×OD+ 1/2 ×AO×OB+ 1/2 ×BO×OC+ 1/2 ×OD×OC= 1/2 ×

AO(OD+OB) + 1/2 ×OC(BO+OD)= 1/2 × (

OD+OB) × (AO+OC)

Hence the area of the rhombus is equal to half of the product of diagonals.

### Examples

**Example 1: Calculate the area of a rhombus whose diagonals are of the length 20 cm and 10 cm?**

**Solution:**

We are Provided that

d1 = 20 cm

d2 = 10 cm

Area of a rhombus, A = (d1 × d2) / 2

= (20 × 10) / 2

= 100 cm

^{2}

Hence, the area of a rhombus is 100 cm^{2}.

**Example 2: Find the area of the rhombus having each side equal to 25 cm and the length of one of its diagonals is equal to 14 cm?**

**Solution:**

ABCD is a rhombus in which AB = BC = CD = DA = 25 cm

AC = 14 cm

Area of rhombus = 1/2 * d

_{1 }* d_{2}

Therefore, BO = 7 cm

In ∆ AOB,

AB² = AO² + OB²

⇒ 25² = 7² + OB²

⇒ 625 = 49 + OB²

⇒ 576 = OB²

⇒ OB = 24

Therefore, BD = 2 x OB = 2 × 24 = 48 cm

Now, area of rhombus = 1/2 × d₁ × d₂ = 1/2 × 14 × 48 = 336 cm²

Hence the area of the rhombus is 336 cm^{2}.

**Example 3: Find the height of the rhombus whose area is 400 m² and the perimeter is 160 m?**

**Solution:**

Given, the perimeter of the rhombus = 160 m

So, side of rhombus = 160/4 = 40 m

We know that the area of any parallelogram = b × h

Therefore the height is:

⇒ 400 = 40 × h

⇒ h = 10 m

Therefore, the height of the rhombus is 10 m.

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