| Content On This Page | ||
|---|---|---|
| Perimeter of Polygons (Triangle, Quadrilateral, etc.) | Perimeter of Squares and Rectangles | Perimeter of Parallelograms |
| Perimeter of Other Simple Plane Figures | ||
Perimeter of Various Plane Figures
Perimeter of Polygons (Triangle, Quadrilateral, etc.)
A polygon is a closed two-dimensional shape made up of a finite number of straight line segments connected end to end. These segments are called the sides of the polygon, and the points where the segments meet are called vertices.
The perimeter of any polygon is simply defined as the sum of the lengths of all its sides. It represents the total distance around the boundary of the polygon.
General Formula for the Perimeter of a Polygon
If a polygon has '$n$' sides with lengths denoted by $s_1, s_2, s_3, ..., s_n$, then its perimeter ($P$) is given by the sum of these lengths:
$\text{Perimeter} (P) = s_1 + s_2 + s_3 + ... + s_n$
... (1)
Using summation notation, this can be written as:
$P = \sum\limits_{i=1}^{n} s_i$
... (2)
Perimeter of Triangles
A triangle is the simplest polygon, having exactly 3 sides. Triangles are classified based on their side lengths and angles.
Let the lengths of the three sides of a triangle be $a$, $b$, and $c$. According to the general formula for polygons, the perimeter ($P_{\text{triangle}}$) is the sum of the lengths of its three sides:
$\text{Perimeter of Triangle} = a + b + c$
... (3)
Special Cases of Triangles:
-
Equilateral Triangle:
An equilateral triangle has all three sides equal in length. Let the length of each side be '$s$'.
Side lengths: $s, s, s$
Using the triangle perimeter formula (3):
$P_{\text{equilateral}} = s + s + s$
$\textbf{Perimeter of Equilateral Triangle} = \mathbf{3s}$
... (3a)
-
Isosceles Triangle:
An isosceles triangle has two sides equal in length. Let the two equal sides have length '$a$', and the third side have length '$b$'.
Side lengths: $a, a, b$
Using the triangle perimeter formula (3):
$P_{\text{isosceles}} = a + a + b$
$\textbf{Perimeter of Isosceles Triangle} = \mathbf{2a + b}$
... (3b)
-
Scalene Triangle:
A scalene triangle has all three sides of different lengths. Let the side lengths be $a, b,$ and $c$, where $a \neq b$, $b \neq c$, and $a \neq c$.
Side lengths: $a, b, c$
Using the triangle perimeter formula (3):
$\textbf{Perimeter of Scalene Triangle} = \mathbf{a + b + c}$
... (3c)
This is the same as the general formula for a triangle, as there is no further simplification possible when all sides are different.
Example 1. Find the perimeter of a triangle with side lengths $5$ cm, $7$ cm, and $9$ cm.
Answer:
Given:
Side lengths of the triangle are $a = 5$ cm, $b = 7$ cm, $c = 9$ cm.
To Find:
Perimeter of the triangle.
Solution:
The perimeter of a triangle is the sum of its side lengths. Using formula (3):
$\text{Perimeter} = a + b + c$
Substitute the given side lengths:
$\text{Perimeter} = 5 \$ \text{cm} + 7 \$ \text{cm} + 9 \$ \text{cm}$
[Substituting values for a, b, c]
$\text{Perimeter} = (5 + 7 + 9) \$ \text{cm}$
$\text{Perimeter} = 21 \$ \text{cm}$
Therefore, the perimeter of the triangle is 21 cm.
Perimeter of Quadrilaterals
A quadrilateral is a polygon with 4 sides. Examples include squares, rectangles, parallelograms, rhombuses, kites, and trapeziums, as well as irregular quadrilaterals.
If the lengths of the four sides of a quadrilateral are $a$, $b$, $c$, and $d$, then its perimeter ($P_{\text{quadrilateral}}$) is the sum of these four lengths, based on the general polygon formula (1):
$\textbf{Perimeter of Quadrilateral} = \mathbf{a + b + c + d}$
... (4)
Specific types of quadrilaterals (like squares, rectangles, parallelograms) have simpler formulas due to properties like equal opposite sides, which are discussed in subsequent sections.
Perimeter of Other Polygons
The principle for finding the perimeter remains consistent for any polygon: simply add the lengths of all its sides.
-
Pentagon (5 sides):
For a pentagon with side lengths $s_1, s_2, s_3, s_4, s_5$, the perimeter is $P = s_1 + s_2 + s_3 + s_4 + s_5$.
If it's a regular pentagon (all 5 sides are equal, say length '$s$'), the perimeter is:
$\textbf{Perimeter of Regular Pentagon} = \mathbf{5s}$
... (5a)
-
Hexagon (6 sides):
For a hexagon with side lengths $s_1, s_2, s_3, s_4, s_5, s_6$, the perimeter is $P = s_1 + s_2 + s_3 + s_4 + s_5 + s_6$.
If it's a regular hexagon (all 6 sides are equal, say length '$s$'), the perimeter is:
$\textbf{Perimeter of Regular Hexagon} = \mathbf{6s}$
... (6a)
-
Regular n-sided Polygon:
If a regular polygon has '$n$' sides, and each side has the same length '$s$', its perimeter is the sum of $n$ times the side length:
$\textbf{Perimeter of Regular n-gon} = \mathbf{n \times s}$
... (N)
Example 2. A rectangular playground has sides of length 80 m, 50 m, 80 m, and 50 m. Find its perimeter.
Answer:
Given:
Side lengths of the quadrilateral (rectangle) are $s_1 = 80$ m, $s_2 = 50$ m, $s_3 = 80$ m, $s_4 = 50$ m.
To Find:
Perimeter of the playground.
Solution:
The perimeter of a polygon is the sum of its side lengths. Using the general formula for a quadrilateral (4):
$\text{Perimeter} = s_1 + s_2 + s_3 + s_4$
Substitute the given side lengths:
$\text{Perimeter} = 80 \$ \text{m} + 50 \$ \text{m} + 80 \$ \text{m} + 50 \$ \text{m}$
$\text{Perimeter} = (80 + 50 + 80 + 50) \$ \text{m}$
$\text{Perimeter} = 260 \$ \text{m}$
Therefore, the perimeter of the rectangular playground is 260 metres.
Example 3. A craftsman is making a decorative border for a hexagonal mirror. If the mirror is a regular hexagon with each side measuring $18$ cm, what length of border material is needed?
Answer:
Given:
Shape is a regular hexagon.
Number of sides ($n$) $= 6$.
Length of each side ($s$) $= 18$ cm.
To Find:
Length of the border material needed, which is the perimeter of the regular hexagon.
Solution:
For a regular polygon with $n$ sides of length $s$, the perimeter is $P = n \times s$. Using formula (N) or (6a):
$\text{Perimeter} = 6 \times s$
Substitute the given side length:
$\text{Perimeter} = 6 \times 18 \$ \text{cm}$
[Substituting $s=18$ cm]
Calculate the product:
$\begin{array}{cc}& & 1 & 8 \\ \times & & & 6 \\ \hline & 1 & 0 & 8 \\ \hline \end{array}$$\text{Perimeter} = 108 \$ \text{cm}$
Therefore, the craftsman needs 108 cm of border material.
Perimeter of Squares and Rectangles
Squares and rectangles are specific types of quadrilaterals (polygons with 4 sides) that are commonly encountered. Their properties of equal opposite sides and right angles lead to simplified formulas for calculating their perimeters.
Perimeter of a Square
A square is defined as a quadrilateral with four equal sides and four right angles ($90^\circ$).
Let the length of each side of the square be '$s$'. Since all four sides are equal, the lengths of the sides are $s, s, s, s$.
Derivation of the Formula:
Using the general formula for the perimeter of a polygon (sum of all sides):
$\text{Perimeter of Square} (P_{\text{square}}) = s + s + s + s$
... (1)
Combine the like terms on the right side:
$\text{Perimeter of Square} = 4s$
... (2)
So, the formula for the perimeter of a square is:
$\textbf{Perimeter of Square} = \mathbf{4 \times (Side \$\$ Length)}$
Or simply, $\mathbf{P = 4s}$.
Example 1. Calculate the perimeter of a square plot of land whose side measures $15$ metres.
Answer:
Given:
Shape is a square.
Side length, $s = 15$ m.
To Find:
Perimeter of the square plot.
Solution:
The formula for the perimeter of a square is $P = 4s$. Using formula (2) derived above:
$\text{Perimeter} = 4 \times s$
Substitute the given side length:
$\text{Perimeter} = 4 \times 15 \$ \text{m}$
[Substituting $s=15$ m]
$\text{Perimeter} = 60 \$ \text{m}$
Therefore, the perimeter of the square plot is 60 metres.
Perimeter of a Rectangle
A rectangle is a quadrilateral with opposite sides equal in length and four right angles ($90^\circ$). The longer side is typically called the length, and the shorter side is called the width or breadth.
Let the length of the rectangle be '$l$' and the width (or breadth) be '$w$'. The sides of the rectangle are $l, w, l, w$.
Derivation of the Formula:
Using the general formula for the perimeter of a polygon (sum of all sides):
$\text{Perimeter of Rectangle} (P_{\text{rectangle}}) = l + w + l + w$
... (3)
Group the like terms together:
$\text{Perimeter of Rectangle} = (l + l) + (w + w)$
... (4)
$\text{Perimeter of Rectangle} = 2l + 2w$
... (5)
Factor out the common factor, which is 2:
$\text{Perimeter of Rectangle} = 2(l + w)$
... (6)
So, the formula for the perimeter of a rectangle is:
$\textbf{Perimeter of Rectangle} = \mathbf{2 \times (Length + Width)}$
Or simply, $\mathbf{P = 2(l+w)}$.
Example 2. A rectangular park is $120$ metres long and $85$ metres wide. What is the length of the wire needed to fence it completely?
Answer:
Given:
Shape is a rectangle.
Length, $l = 120$ m.
Width, $w = 85$ m.
To Find:
The total length of wire needed, which is the perimeter of the park.
Solution:
The formula for the perimeter of a rectangle is $P = 2(l + w)$. Using formula (6) derived above:
$\text{Perimeter} = 2(l + w)$
Substitute the given length and width:
$\text{Perimeter} = 2 \times (120 \$ \text{m} + 85 \$ \text{m})$
[Substituting $l=120$ m and $w=85$ m]
$\text{Perimeter} = 2 \times (205 \$ \text{m})$
$\text{Perimeter} = 410 \$ \text{m}$
Therefore, the length of the wire needed to fence the park is 410 metres.
Perimeter of Parallelograms
A parallelogram is a special type of quadrilateral where both pairs of opposite sides are parallel and equal in length. This property simplifies the calculation of its perimeter.
Consider a parallelogram ABCD. Let the length of side AB be '$a$' and the length of side BC be '$b$'. Since opposite sides are equal in length, we have:
AB = CD = $a$
BC = DA = $b$
Formula and Derivation
The perimeter of any polygon is the sum of the lengths of all its sides. For parallelogram ABCD, the sides are AB, BC, CD, and DA.
Using the general formula for polygon perimeter:
$\text{Perimeter of Parallelogram} (P_{\text{parallelogram}}) = \text{AB} + \text{BC} + \text{CD} + \text{DA}$
... (1)
Substitute the side lengths $a$ and $b$ into the formula:
$\text{Perimeter of Parallelogram} = a + b + a + b$
[Substituting side lengths in (1)]
Now, group the like terms:
$\text{Perimeter of Parallelogram} = (a + a) + (b + b)$
... (2)
$\text{Perimeter of Parallelogram} = 2a + 2b$
... (3)
Finally, factor out the common factor of 2:
$\text{Perimeter of Parallelogram} = 2(a + b)$
... (4)
So, the formula for the perimeter of a parallelogram is:
$\textbf{Perimeter of Parallelogram} = \mathbf{2 \times (Sum \$\$ of \$\$ lengths \$\$ of \$\$ adjacent \$\$ sides)}$
Or simply, $\mathbf{P = 2(a+b)}$, where $a$ and $b$ are the lengths of two adjacent sides.
It is worth noting that this formula is identical to that of a rectangle ($P=2(l+w)$). This is because a rectangle is a special type of parallelogram where all four angles are $90^\circ$, but it still maintains the property of opposite sides being equal.
Special Case: Rhombus
A rhombus is a parallelogram in which all four sides are equal in length. It's a special case of a parallelogram where the adjacent sides are equal ($a=b$).
Let the length of each side of the rhombus be '$s$'. Since all four sides are equal, the side lengths are $s, s, s, s$.
Derivation of the Formula:
Using the general formula for the perimeter of a polygon (sum of all sides):
$\text{Perimeter of Rhombus} (P_{\text{rhombus}}) = s + s + s + s$
... (5)
Combine the like terms:
$\textbf{Perimeter of Rhombus} = \mathbf{4s}$
... (6)
Alternatively, we can use the parallelogram formula $P=2(a+b)$. For a rhombus, $a=s$ and $b=s$. Substituting these into the parallelogram formula:
$P_{\text{rhombus}} = 2(s + s) = 2(2s) = 4s$
[Using formula (4) with a=s, b=s]
Both methods yield the same formula:
$\textbf{Perimeter of Rhombus} = \mathbf{4 \times (Side \$\$ Length)}$
Or simply, $\mathbf{P = 4s}$.
This formula is identical to the formula for a square ($P=4s$). This is because a square is a special type of rhombus where all four angles are $90^\circ$, but it still maintains the property of all sides being equal.
Examples
Example 1. Find the perimeter of a parallelogram whose adjacent sides measure $8$ cm and $6$ cm.
Answer:
Given:
Shape is a parallelogram.
Length of one adjacent side, $a = 8$ cm.
Length of the other adjacent side, $b = 6$ cm.
To Find:
Perimeter of the parallelogram.
Solution:
The formula for the perimeter of a parallelogram is $P = 2(a + b)$. Using formula (4) derived above:
$\text{Perimeter} = 2(a + b)$
Substitute the given lengths of the adjacent sides:
$\text{Perimeter} = 2 \times (8 \$ \text{cm} + 6 \$ \text{cm})$
[Substituting $a=8$ cm and $b=6$ cm]
$\text{Perimeter} = 2 \times (14 \$ \text{cm})$
$\text{Perimeter} = 28 \$ \text{cm}$
Therefore, the perimeter of the parallelogram is 28 cm.
Example 2. A field is in the shape of a rhombus. If the length of its side is $50$ m, find the perimeter of the field.
Answer:
Given:
Shape is a rhombus.
Side length, $s = 50$ m.
To Find:
Perimeter of the rhombus field.
Solution:
The formula for the perimeter of a rhombus is $P = 4s$. Using formula (6) derived above:
$\text{Perimeter} = 4 \times s$
Substitute the given side length:
$\text{Perimeter} = 4 \times 50 \$ \text{m}$
[Substituting $s=50$ m]
$\text{Perimeter} = 200 \$ \text{m}$
Therefore, the perimeter of the field is 200 metres.
Alternate Solution:
A rhombus is a parallelogram where adjacent sides are equal. Using the parallelogram formula $P=2(a+b)$ with $a=50$ m and $b=50$ m:
$\text{Perimeter} = 2(50 \$ \text{m} + 50 \$ \text{m})$
[Using formula (4)]
$\text{Perimeter} = 2 \times 100 \$ \text{m}$
$\text{Perimeter} = 200 \$ \text{m}$
Both methods give the same result.
Perimeter of Other Simple Plane Figures
While polygons with straight sides have their perimeter calculated by summing side lengths, other plane figures may involve curved boundaries, requiring different formulas. Here, we look at a few other common shapes, including some quadrilaterals and circles/semicircles.
Perimeter of a Trapezium
A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides. Unlike parallelograms, the non-parallel sides may or may not be equal.
Let the lengths of the four sides of a general trapezium be $a$, $b$, $c$, and $d$. There is no general simplification for the perimeter of *any* trapezium.
Formula:
The perimeter is simply the sum of the lengths of all four sides:
$\textbf{Perimeter of Trapezium} = \mathbf{a + b + c + d}$
Special Case: Isosceles Trapezium
An isosceles trapezium is a trapezium where the non-parallel sides are equal in length. Let the parallel sides be $a$ and $c$, and the equal non-parallel sides be $b$.
Using the general trapezium formula with side lengths $a, b, c, b$:
$\text{Perimeter of Isosceles Trapezium} = a + b + c + b$
Combine the like terms:
$\textbf{Perimeter of Isosceles Trapezium} = \mathbf{a + c + 2b}$
Perimeter of a Kite
A kite is a quadrilateral that has two distinct pairs of equal-length adjacent sides. It is not a parallelogram unless it is also a rhombus.
Let the lengths of the two distinct adjacent sides be '$a$' and '$b$'. Since there are two pairs of equal adjacent sides, the side lengths of the kite are $a, b, b, a$ (or $a, a, b, b$).
Derivation of the Formula:
Using the general formula for the perimeter of a polygon (sum of all sides):
$\text{Perimeter of Kite} (P_{\text{kite}}) = a + a + b + b$
... (1)
Combine the like terms:
$\text{Perimeter of Kite} = 2a + 2b$
... (2)
Factor out the common factor 2:
$\textbf{Perimeter of Kite} = \mathbf{2(a + b)}$
... (3)
So, the formula for the perimeter of a kite is:
$\textbf{Perimeter of Kite} = \mathbf{2 \times (Sum \$\$ of \$\$ lengths \$\$ of \$\$ the \$\$ two \$\$ distinct \$\$ adjacent \$\$ sides)}$
Or simply, $\mathbf{P = 2(a+b)}$.
Perimeter of a Circle (Circumference)
The perimeter of a circle has a special name: Circumference. Unlike polygons, a circle has a smooth, curved boundary.
The circumference of a circle is directly proportional to its radius or diameter. The constant of proportionality is the mathematical constant $\pi$ (pi).
The value of $\pi$ is an irrational number, approximately equal to $3.14159$. For many calculations, approximations like $\frac{22}{7}$ or $3.14$ are used.
Let $r$ be the radius of the circle and $d$ be its diameter. Remember that the diameter is twice the radius, so $d = 2r$.
Formula:
The formula for the circumference (C) of a circle is:
$\textbf{Circumference} (C) = \mathbf{2 \pi r}$
... (4)
Alternatively, since $d = 2r$, we can substitute $2r$ with $d$ in formula (4):
$C = \pi \times (2r) = \pi d$
$\textbf{Circumference} (C) = \mathbf{\pi d}$
... (5)
So, the circumference can be calculated using either the radius or the diameter:
$\textbf{Circumference of Circle} = \mathbf{2 \times \pi \times Radius} = \mathbf{\pi \times Diameter}$
Perimeter of a Semicircle
A semicircle is half of a circle, cut along its diameter. Its boundary consists of a curved arc and a straight line segment (the diameter).
Let $r$ be the radius of the semicircle.
Derivation of the Formula:
The perimeter of a semicircle is the sum of the length of its curved arc and the length of its diameter.
- The length of the curved arc is half the circumference of the full circle with the same radius.
Length of arc $= \frac{1}{2} \times (\text{Circumference of full circle})$
Length of arc $= \frac{1}{2} \times (2 \pi r)$
[Using formula (4)]
Length of arc $= \pi r$
- The length of the straight edge is the diameter of the semicircle, which is $2r$.
Length of diameter $= 2r$
The total perimeter of the semicircle is the sum of these two lengths:
$\text{Perimeter of Semicircle} (P_{\text{semicircle}}) = (\text{Length of arc}) + (\text{Length of diameter})$
... (6)
$\text{Perimeter of Semicircle} = \pi r + 2r$
[Substituting values into (6)]
Factor out the common factor $r$:
$\textbf{Perimeter of Semicircle} = \mathbf{r(\pi + 2)}$
... (7)
So, the formula for the perimeter of a semicircle is:
$\textbf{Perimeter of Semicircle} = \mathbf{Radius \times (\pi + 2)}$
Examples
Example 1. Find the perimeter of a trapezium whose sides are $10$ cm, $8$ cm, $6$ cm, and $5$ cm.
Answer:
Given:
Shape is a trapezium.
Side lengths are $a = 10$ cm, $b = 8$ cm, $c = 6$ cm, $d = 5$ cm.
To Find:
Perimeter of the trapezium.
Solution:
The perimeter of any polygon is the sum of the lengths of all its sides. For a trapezium with sides $a, b, c, d$, the perimeter is $P = a + b + c + d$.
$\text{Perimeter} = 10 \$ \text{cm} + 8 \$ \text{cm} + 6 \$ \text{cm} + 5 \$ \text{cm}$
[Summing all side lengths]
$\text{Perimeter} = (10 + 8 + 6 + 5) \$ \text{cm}$
$\text{Perimeter} = 29 \$ \text{cm}$
The perimeter of the trapezium is 29 cm.
Example 2. Find the circumference of a circular disc with a radius of $7$ cm. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Shape is a circle.
Radius, $r = 7$ cm.
Value of $\pi = \frac{22}{7}$.
To Find:
Circumference of the circular disc.
Solution:
The formula for the circumference of a circle using radius is $C = 2 \pi r$. Using formula (4) derived above:
$C = 2 \pi r$
Substitute the given values:
$C = 2 \times \frac{22}{7} \times 7 \$ \text{cm}$
[Substituting $\pi = \frac{22}{7}$ and $r=7$ cm]
Cancel out the common factor 7 in the numerator and denominator:
$C = 2 \times \frac{22}{\cancel{7}} \times \cancel{7} \$ \text{cm}$
$C = 2 \times 22 \$ \text{cm}$
$C = 44 \$ \text{cm}$
Therefore, the circumference of the circular disc is 44 cm.
Example 3. Calculate the perimeter of a semicircular protractor whose radius is $14$ cm. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Shape is a semicircle.
Radius, $r = 14$ cm.
Value of $\pi = \frac{22}{7}$.
To Find:
Perimeter of the semicircular protractor.
Solution:
The perimeter of a semicircle consists of the curved arc and the diameter. The formula is $P = r(\pi + 2)$. Using formula (7) derived above:
$\text{Perimeter} = r(\pi + 2)$
Substitute the given values:
$\text{Perimeter} = 14 \$ \text{cm} \times \left(\frac{22}{7} + 2\right)$
[Substituting $r=14$ cm and $\pi = \frac{22}{7}$]
First, calculate the term inside the parenthesis by finding a common denominator:
$\frac{22}{7} + 2 = \frac{22}{7} + \frac{2 \times 7}{1 \times 7} = \frac{22}{7} + \frac{14}{7}$
$= \frac{22 + 14}{7} = \frac{36}{7}$
Now substitute this sum back into the perimeter formula:
$\text{Perimeter} = 14 \$ \text{cm} \times \frac{36}{7}$
Cancel out the common factor 7:
$\text{Perimeter} = \cancel{14}^2 \$ \text{cm} \times \frac{36}{\cancel{7}_1}$
$\text{Perimeter} = 2 \times 36 \$ \text{cm}$
$\text{Perimeter} = 72 \$ \text{cm}$
Therefore, the perimeter of the semicircular protractor is 72 cm.
Alternate Solution:
Calculate the length of the curved arc and the diameter length separately, and then add them.
Length of curved arc $= \pi r = \frac{22}{7} \times 14 \$ \text{cm}$
Length of arc $= \frac{22}{\cancel{7}_1} \times \cancel{14}^2 \$ \text{cm} = 22 \times 2 \$ \text{cm} = 44 \$ \text{cm}$
Diameter length $= 2r = 2 \times 14 \$ \text{cm} = 28 \$ \text{cm}$.
Perimeter = Arc length + Diameter length
$\text{Perimeter} = 44 \$ \text{cm} + 28 \$ \text{cm}$
$\text{Perimeter} = 72 \$ \text{cm}$
Both methods yield the same result.