Perimeter of Rhombus Formula: Derivation & Properties

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The perimeter of a Rhombus formula is 4 x s (Here, s = the side length of the rhombus). Rhombus is a flat-shaped two dimensional structure having four equal angles and four equal sides. The four angles are not 90 degrees necessarily. 

This shape is also considered to be a diamond-shaped structure. The perimeter of the rhombus is actually the total distance covered along the border of it. 

Key Terms: Rhombus, Perimeter, Pythagoras Theorem, Adjacent Angles, Bisect, Diagonals


Perimeter of Rhombus

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According to the definition of the Rhombus, 

Perimeter of Rhombus = 4 x s

Where,

s = the side length of the rhombus.

Hence, P = s + s + s + s = 4s

Rhombus

Rhombus


Area of the Rhombus

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The area of rhombus can be determined in two ways:

i) By multiplying the height & base of the rhombus

Area = b x h [b = base of rhombus; h = height of rhombus]

ii) By determining the product of the diagonal of the rhombus & dividing it by 2

Area = ½ x d1 x d2 [d1 & d2 are the diagonals of the rhombus]

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Derivation of the Area of Rhombus

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If MNOP is considered a Rhombus with base MN = b, PN ⊥ MO, MO is a diagonal of rhombus = d2; PN is a diagonal of rhombus = d1; altitude from O on MN is OZ which is = h.

  • Then, the area of the rhombus MNOP = 2 x area of ∆ MNO

= 2 x ½ MN x OP

= 2 x ½ b x h (sq. Units)

  • The area of the rhombus = 4 x Area of ∆ MZP

= 4 x ½ x MZ x ZN 

= 4 x ½ x d2 x ½ d1 (sq. Units)

= 4 x 1/8 d1 x d2

The area of a rhombus = ½ x d1 x d2

Finding Perimeter of Rhombus Using Diagonal Lengths

Given, in a rhombus PQRS, a as the horizontal diagonal length and b as vertical diagonal length and center o.

Let’s take triangle POS, then the length of
OP = a/2

OS = b/2

If we assume the side of the rhombus as = X

Using Pythagoras theorem, 

X2 = a2/2 + b2/2

X2 = (a2 + b2)/4

X = \(\sqrt{ \frac{( a^2+b^2)}{2}}\)

By substituting the value of X in the rhombus formula = 4X, we get, 

4 x \(\sqrt{ \frac{( a^2+b^2)}{2}}\)

So, the perimeter will be = 2 x \(\sqrt{ \frac{( a^2+b^2)}{2}}\)


Properties of Rhombus

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Rhombus is known to be a square due to the following properties of Rhombus:

  • It’s all sides have equal length
  • It’s opposite sides are parallel
  • Its diagonals bisect each other at 90 degrees.
  • Two adjacent angles forms a total of 180 degree
  • It’s altitude is the distance at right angles between the two parallel sides

Things to Remember

  • Diagonals of a rhombus divide each other at right angles or 90 degrees.
  • The perimeter of a rhombus is said to be the sum of all four sides
  • The sum of all the angles of a rhombus is 360 degrees
  • The rhombus share the same properties as that of a square. Hence, rhombus is also said to be a square.
  • Rhombus differs from rectangle in a way as rhombus has opposites angles as equivalent while rectangle has opposite sides as equivalent

Sample Questions

Ques. What is the perimeter of a rhombus with side 8 cm? (2 marks)

Ans: Given, side of the Rhombus = 8 cm

Perimeter of rhombus = 4 x s

= 4 x 8 cm = 32 cm 

Ques. The diagonals of a rhombus are 44 cm and 15 cm. Find the area of the rhombus. (2 marks)

Ans: Let’s take, a = 44 cm; b = 15 cm 

Area of the rhombus (A) = ½ x a x b

= ½ x 44 cm x 15 cm

= 330 cm2

Ques. The side of a rhombus is 2.5 cm. What is the perimeter of the rhombus? (2 marks)

Ans: Let’s take, s = 2.5 cm

Perimeter of rhombus (given) = 4 x s

Hence, (P) = 4 x 2.5 cm

= 10 cm

Ques. The diagonals of a rhombus are 24 cm and 10 cm. Find the area of the rhombus. (2 marks)

Ans: Let’s take, a = 24 cm; b = 10 cm 

Area of the rhombus (A) = ½ x a x b 

= ½ x 24 cm x 10 cm

= 60 cm2

Ques. What is the perimeter of a rhombus with side 16 cm? (2 marks)

Ans: Given, side of the Rhombus, s = 16 cm

Perimeter of rhombus = 4 x s

= 4 x 16 cm = 64 cm 

Ques. What is the perimeter of a rhombus having two diagonals of 6cm & 8 cm respectively? (3 marks)

Ans: Given, d1 = 6 cm; d2 = 8 cm

According to the formula, 

Perimeter of rhombus = 2 x \(\sqrt{ \frac{( a^2+b^2)}{2}}\)

= 2 x \(\sqrt{ \frac{( 6^2+8^2}{2}}\)

= 2 x 10 

= 20 cm

Ques. Find the perimeter of a rhombus having two diagonals of 4 cm & 6 cm respectively? (3 marks)

Ans: Given, d1 = 4 cm; d2 = 6 cm

According to the formula, 

Perimeter of rhombus (P) = 2 x \(\sqrt{ \frac{( a^2+b^2)}{2}}\)

= 2 x \(\sqrt{ \frac{( 4^2+6^2}{2}}\)

= 2 x 5.8 cm 

= 11.6 cm

Ques. The Perimeter of a rhombus is 48 cm. Find the side of a rhombus. (3 marks)

Ans: Given, perimeter of the Rhombus (P) = 48cm 

Hence, P = 4 x s [according to formula]

s = p/4

s = 48/4 cm = 12 cm

So, the side of the rhombus is 12 cm

Ques. Find the perimeter of a rhombus with a given side of 5 cm. (3 marks)

Ans: Given, side of a rhombus, s = 5 cm

According to the formula, perimeter of the rhombus is

P = 4 x s

= 4 x 5 cm

= 20 cm

Ques. The perimeter of a rhombus is 60 cm. Find the side of a rhombus. (3 marks)

Ans: Given, perimeter of the Rhombus (P) = 60 cm 

Hence, Perimeter of the rhombus, P = 4 x s [according to the formula]

s = p/4

s = 60/4 cm = 15 cm

So, the side of the rhombus is 15 cm

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CBSE X Related Questions

  • 1.
    Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)


      • 2.
        There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

          • 144
          • 2
          • 420
          • 272

        • 3.
          In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.


            • 4.
              PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.


                • 5.
                  In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).


                    • 6.
                      In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

                        • \( 130^\circ \)
                        • \( 50^\circ \)
                        • \( 120^\circ \)
                        • \( 90^\circ \)

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