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.
    If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

      • \(-2\)
      • \(\sqrt{\frac{8}{3}}\)
      • \(1\)
      • \(2\)

    • 2.
      Which of the following graphs represents a polynomial with both zeroes being positive?


      • 3.
        For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

          • \(0\)
          • \(\frac{80}{81}\)
          • \(1\)
          • \(\frac{82}{80}\)

        • 4.
          It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

            • \(\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle EDF} = \frac{AB}{ED}\)
            • \(\frac{AB}{ED} = \frac{AC}{EF}\)
            • \(\angle A = \angle D, \angle C = \angle F\)
            • \(\frac{AB + BC}{AC} = \frac{ED + DF}{EF}\)

          • 5.
            If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

              • \( -\frac{2}{3} \)
              • \( \frac{2}{3} \)
              • \( \frac{1}{3} \)
              • \( -\frac{1}{3} \)

            • 6.
              The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

                • \((x + 2), (x - 5)\)
                • \((x + 5), (x - 2)\)
                • \((x + 1), (x - 5)\)
                • \((x - 1), (x + 2)\)

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