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Last modified on August 3rd, 2023

A parallelogram has some properties similar to that of a rectangle. We also consider a parallelogram as a slanting rectangle. However, a parallelogram is slightly different from a rectangle in terms of some properties. So its perimeter is found differently. Let us learn how to find the perimeter of a parallelogram here.

The perimeter of a parallelogram is the total distance covered around its edge. In other words, the perimeter of a parallelogram is the total sum of its four sides.

Since the perimeter measures length or distance, its unit is always linear. e.g., as cm, m, inch, ft, or yd.

There are different ways to find the perimeter of a parallelogram based on the information given.

Let ‘a’ and ‘b’ be the sides of a parallelogram. We know that the perimeter of a parallelogram is the sum of all its sides. Also, the opposite sides of a parallelogram are parallel and equal to each other. Thus, the perimeter ‘P’ of a parallelogram is:

P = a + b + a + b units

P = 2a + 2b

P = 2(a + b)

Thus, the standard formula to calculate the perimeter of a parallelogram is given below:

Let us solve an example to clear your concept better.

**Find the perimeter of a parallelogram whose base and side lengths are 6 cm and 9 cm, respectively.**

Solution:

As we know,

P = 2(a + b), here a = 6 cm, b = 9 cm

= 2(6 + 9)

= 2 × 15

= 30 cm

We might not always know all the sides of a parallelogram. The perimeter of a parallelogram can be calculated when other information is given, as in the following cases –

- When one side and diagonals are given
- When base, height, and any angle are given

Consider a parallelogram ABCD with sides ‘a’ & ‘b’ and diagonals ‘x’ & ‘y’. Here, ‘a’, ‘x’ and ‘y’ are known. The value of ‘b’ is unknown. We have to calculate the perimeter of the parallelogram.

**Derivation**

Applying the law of cosines for the ΔABC,

x^{2} = a^{2} + b^{2} – 2ab cos∠ ABC

Applying the cosine rule for the ΔBAD

y^{2} = a^{2} + b^{2} – 2ab cos∠ BAD

Adding the above two equations,

x^{2} + y^{2} = 2a^{2} + 2b^{2} – 2ab (cos∠ABC + cos∠BAD) …. (1)

Since, any 2 adjacent angles of a parallelogram add up to 180° (supplementary angles), so,

∠ABC + ∠BAD = 180°

or, ∠ABC = 180° – ∠BAD

Applying cos on both sides,

cos∠ABC = cos(180° – ∠BAD) = – cos∠BAD

⇒ x^{2} + y^{2} = 2a^{2} + 2b^{2} – 2ab ( – cos∠BAD + cos∠BAD) , from …. (1)

⇒ x^{2} + y^{2} = 2a^{2} + 2b^{2} – 2ab (0)

⇒ x^{2} + y^{2} = 2a^{2} + 2b^{2} …. (the relation between the sides and diagonals)

⇒ 2b^{2} = x^{2} + y^{2} – 2a^{2}

⇒ b^{2} = (x^{2} + y^{2} – 2a^{2})/2

⇒ b = √[(x^{2} + y^{2} – 2a^{2})/2]

as we know,

P = 2(a + b)

⇒ P = 2a + 2 √[(x^{2} + y^{2} – 2a^{2})/2]

⇒ P = 2a + √[2(x^{2} + y^{2} – 2a^{2})

**⇒ P = 2a + √(2x ^{2} + 2y^{2} – 4a^{2})**

The formula to calculate the perimeter of a parallelogram with one side and the 2 diagonals is given below:

Finding the perimeter of a parallelogram when **ONE SIDE** and **DIAGONALS** are known

**Find the perimeter of a parallelogram with the side length 7 cm and diagonals 9 cm and 11 cm. Round your answer to two decimals.**

Solution:

As we know,

P = 2a + √(2x^{2} + 2y^{2} – 4a^{2}) , here a = 7 cm, x = 9 cm, y = 11 cm

= 2 × 7 + √(2 × 9^{2} + 2 × 11^{2} – 4 × 7^{2})

= 14 + √(162 + 242 – 196)

= 14 + √208

= 28.42 cm

Let us find the perimeter of parallelogram ABCD with base, height and any vertex angle.

**Derivation**

In ΔCED,

sinθ = h/a

a = h/sinθ

∴The perimeter (P) of the parallelogram is,

P = 2a + 2b

⇒ P = 2(h/sinθ) + 2b

Here θ any vertex angle because any two adjacent angles of a parallelogram are supplementary. So,

**sinθ = sin(180° – θ),** for any θ

The formula to calculate the perimeter of a parallelogram with base, height, and any vertex angle is given below:

Let us solve an example to understand the above formula better.

Finding the perimeter of a parallelogram when** BASE, HEIGHT,** and **VERTEX ANGLE** are known

**Find the perimeter of a parallelogram where one of its sides is 12 ft, its corresponding height is 8 ft, and one of the vertex angles is 30 degrees.**

Solution:

As we know,

P = 2(h/sinθ) + 2b , here h = 8 ft, b = 12 ft, θ = 30°

= 2(8/sin 30°) + 2 × 12

= 2(8 ÷ 1/2) + 24 , since sin 30° = 1/2

= 32 + 24

= 56 ft

Let us find the relationship between the area and perimeter of a parallelogram.

We know,

Area of a parallelogram (A) = b x h square units ……(1), here b = base, h = height

And,

Perimeter of a parallelogram (P) = 2(a + b) units, here a & b are any 2 adjacent sides

Now, value of b in terms of P is

⇒ P/2 = a + b

b = (P/2) – a

A = ((P/2) – a)h, Substitute the value of b from (1)

∴**A = ((P/2) – a)h** square units

Let us solve some examples to understand the relationship.

Finding the perimeter of a parallelogram when **ONE SIDE**, **HEIGHT,** and **AREA** are known

**Find the perimeter of a parallelogram with a height of 6 cm and an adjacent side of 8 (which is not the base) cm whose area is 36 sq. cm.**

Solution:

As we know,

A = ((P/2) – a)h, here a = 8 cm, h = 6 cm, A = 36 sq. cm

∴ P = 2((A/h) + a)

= 2(36/6 + 8)

= 28 cm

Finding the length of one side of a parallelogram when **BASE** and **PERIMETER** are known

**Find the length of another side of the parallelogram whose base is 6 cm and the perimeter is 42 cm.**

Solution:

As we know,

a = P/2 – b , here a = unknown side, b = 6 cm, P = 42 cm

= 42/2 – 6

= 21 – 6 = 15 cm

Last modified on August 3rd, 2023