Basic MathGuides

Perimeter

Understanding Perimeter

The perimeter of a shape is the total distance around its boundary. It is the sum of all the sides of a shape. Perimeter is measured in linear units such as meters, centimeters, feet, inches, etc.

Regular Polygons

1. Square

s
s
s
s

Formula: P = 4s

where s = length of one side

Example: If a square has sides of 5 cm, the perimeter is:

P = 4 × 5 cm = 20 cm

2. Rectangle

l
w
l
w

Formula: P = 2(l + w)

where l = length and w = width

Example: If a rectangle has length 8 cm and width 5 cm, the perimeter is:

P = 2 × (8 cm + 5 cm) = 2 × 13 cm = 26 cm

3. Regular Triangle (Equilateral)

s
s
s

Formula: P = 3s

where s = length of one side

Example: If an equilateral triangle has sides of 7 cm, the perimeter is:

P = 3 × 7 cm = 21 cm

4. Regular Pentagon

s s s s s

Formula: P = 5s

where s = length of one side

Example: If a regular pentagon has sides of 6 cm, the perimeter is:

P = 5 × 6 cm = 30 cm

5. Regular n-sided Polygon

s s s s s s s s

Formula: P = n × s

where n = number of sides and s = length of one side

Example: If a regular octagon (8 sides) has sides of 4 cm, the perimeter is:

P = 8 × 4 cm = 32 cm

Circles

Circle (Circumference)

r

Formula: C = 2πr = πd

where r = radius, d = diameter, and π ≈ 3.14159

Example: If a circle has radius 7 cm, the circumference is:

C = 2 × π × 7 cm = 14π cm ≈ 43.98 cm

Alternative calculation: If the diameter is 14 cm:

C = π × 14 cm = 14π cm ≈ 43.98 cm

Irregular Shapes

1. Irregular Polygon

a b c d e f

Formula: P = sum of all sides

P = a + b + c + d + e + f + ...

Example: If an irregular hexagon has sides of 8 cm, 6 cm, 7 cm, 9 cm, 5 cm, and 7 cm, the perimeter is:

P = 8 cm + 6 cm + 7 cm + 9 cm + 5 cm + 7 cm = 42 cm

2. Composite Shapes

a b c d

Method: Identify all outer edges of the shape

P = sum of all outer edges

Example: For a shape composed of a rectangle with a semicircle:

P = 2 × length + width + semicircle arc

If the rectangle is 8 cm × 5 cm and the semicircle has diameter 5 cm:

P = 8 cm + 5 cm + 8 cm + (π × 5 cm)/2 = 21 cm + 7.85 cm = 28.85 cm

Special Cases and Applications

1. Perimeter of a Triangle Using Coordinates

When you have the vertices (coordinates) of a triangle:

Given points A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃):

Side AB = √[(x₂ - x₁)² + (y₂ - y₁)²]

Side BC = √[(x₃ - x₂)² + (y₃ - y₂)²]

Side CA = √[(x₁ - x₃)² + (y₁ - y₃)²]

Perimeter = AB + BC + CA

Example: For a triangle with vertices at (0,0), (4,0), and (2,3):

AB = √[(4 - 0)² + (0 - 0)²] = 4

BC = √[(2 - 4)² + (3 - 0)²] = √(4 + 9) = √13 ≈ 3.61

CA = √[(0 - 2)² + (0 - 3)²] = √(4 + 9) = √13 ≈ 3.61

Perimeter = 4 + 3.61 + 3.61 = 11.22 units

2. Perimeter of a Sector

r arc r

Formula: P = 2r + arc length

Where arc length = r × θ (θ in radians)

Example: For a sector with radius 10 cm and central angle 60° (π/3 radians):

P = 2 × 10 cm + 10 cm × (π/3) = 20 cm + 10.47 cm = 30.47 cm

3. Perimeter with Missing Information

When some measurements are unknown, you can use geometric properties:

  • Pythagorean Theorem: In a right triangle, if legs are a and b, and hypotenuse is c, then c² = a² + b²
  • Properties of Special Triangles:
    • Isosceles triangle: two sides are equal
    • 30-60-90 triangle: sides are in the ratio 1 : √3 : 2
    • 45-45-90 triangle: sides are in the ratio 1 : 1 : √2
  • Properties of Parallelograms: Opposite sides are equal

Real-World Applications

1. Fencing a Garden

To calculate how much fencing is needed for a garden, you need to find the perimeter.

Example: A rectangular garden is 15 m long and 8 m wide. How much fencing is needed?

P = 2(l + w) = 2(15 m + 8 m) = 2(23 m) = 46 m of fencing

2. Picture Frame

To find how much framing material is needed, calculate the perimeter.

Example: A rectangular picture is 24 cm × 18 cm. How much frame molding is needed?

P = 2(l + w) = 2(24 cm + 18 cm) = 2(42 cm) = 84 cm of molding

If we include a 10% waste allowance: 84 cm + 8.4 cm = 92.4 cm

3. Track Length

A running track typically consists of two straight sections and two semicircular ends.

Example: A track has straight sections of 100 m each and semicircular ends with radius 30 m. What is the length of one lap?

P = 2 × straight sections + 2 × semicircle perimeters

P = 2 × 100 m + 2 × (π × 30 m) = 200 m + 188.5 m = 388.5 m

Perimeter Quiz

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