Perimeter and Area - Grade 8

1. Perimeter

Definition: Perimeter is the total distance around the boundary of a two-dimensional shape.

Perimeter Formulas:

Shape	Perimeter Formula	Variables
Square	$ P = 4s $	s = side length
Rectangle	$ P = 2(l + w) $	l = length, w = width
Triangle	$ P = a + b + c $	a, b, c = side lengths
Parallelogram	$ P = 2(a + b) $	a, b = side lengths
Trapezoid	$ P = a + b + c + d $	a, b, c, d = all sides
Regular Polygon	$ P = n \times s $	n = number of sides, s = side length

Key Points:

Perimeter is always measured in linear units (cm, m, ft, etc.)
Add all side lengths to find perimeter
For regular polygons, multiply one side by the number of sides

Examples:

Example 1: Find the perimeter of a rectangle with length 12 cm and width 5 cm.

$ P = 2(l + w) = 2(12 + 5) = 2(17) = 34 $ cm

Example 2: Find the perimeter of a triangle with sides 8 m, 6 m, and 10 m.

$ P = 8 + 6 + 10 = 24 $ m

2. Area

Definition: Area is the amount of space inside the boundary of a two-dimensional shape.

Area Formulas:

Shape	Area Formula	Variables
Square	$ A = s^2 $	s = side length
Rectangle	$ A = l \times w $	l = length, w = width
Triangle	$ A = \frac{1}{2}bh $	b = base, h = height
Parallelogram	$ A = b \times h $	b = base, h = height
Trapezoid	$ A = \frac{1}{2}(b_1 + b_2)h $	b₁, b₂ = bases, h = height
Rhombus	$ A = \frac{1}{2}d_1 d_2 $	d₁, d₂ = diagonals

Key Points:

Area is always measured in square units (cm², m², ft², etc.)
Height must be perpendicular to the base
Different shapes have different area formulas

Examples:

Example 1: Find the area of a rectangle with length 15 cm and width 8 cm.

$ A = l \times w = 15 \times 8 = 120 $ cm²

Example 2: Find the area of a triangle with base 10 m and height 6 m.

$ A = \frac{1}{2}bh = \frac{1}{2}(10)(6) = 30 $ m²

3. Area and Perimeter: Word Problems

Steps to Solve Word Problems:

Read the problem carefully and identify what is given
Determine what you need to find (area or perimeter)
Draw a diagram if helpful
Choose the correct formula
Substitute values and solve
Include units in your answer

Examples:

Example 1: A rectangular garden is 20 feet long and 12 feet wide. How much fencing is needed to enclose the garden?

Looking for: Perimeter (fencing around the garden)

$ P = 2(l + w) = 2(20 + 12) = 2(32) = 64 $ feet

Example 2: A square patio has sides of 8 meters. How many square meters of tile are needed to cover the patio?

Looking for: Area (covering the surface)

$ A = s^2 = 8^2 = 64 $ m²

Example 3: A rectangular room is 15 feet by 10 feet. Carpeting costs $3 per square foot. What is the total cost to carpet the room?

Step 1: Find area: $ A = 15 \times 10 = 150 $ ft²

Step 2: Calculate cost: $ 150 \times 3 = \$450 $

Example 4: A triangular sail has a base of 6 meters and a height of 8 meters. What is its area?

$ A = \frac{1}{2}bh = \frac{1}{2}(6)(8) = 24 $ m²

4. Area and Circumference of Circles

Key Terms:

Radius (r): Distance from center to edge
Diameter (d): Distance across circle through center; $ d = 2r $
Circumference (C): Perimeter/distance around the circle
Pi (π): Approximately 3.14 or $ \frac{22}{7} $

Formulas:

Circumference: $ C = 2\pi r $ or $ C = \pi d $

Area: $ A = \pi r^2 $

Examples:

Example 1: Find the circumference and area of a circle with radius 7 cm. (Use $ \pi = \frac{22}{7} $)

Circumference: $ C = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 $ cm

Area: $ A = \pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154 $ cm²

Example 2: A circular pool has a diameter of 12 meters. Find its area. (Use $ \pi = 3.14 $)

First find radius: $ r = \frac{d}{2} = \frac{12}{2} = 6 $ m

Area: $ A = \pi r^2 = 3.14 \times 6^2 = 3.14 \times 36 = 113.04 $ m²

Example 3: Find the radius of a circle with circumference 31.4 cm. (Use $ \pi = 3.14 $)

$ C = 2\pi r $

$ 31.4 = 2 \times 3.14 \times r $

$ 31.4 = 6.28r $

$ r = \frac{31.4}{6.28} = 5 $ cm

5. Area and Perimeter of Semicircles and Quarter Circles

Semicircle (Half Circle):

Area of Semicircle: $ A = \frac{1}{2}\pi r^2 $ or $ A = \frac{\pi r^2}{2} $

Perimeter of Semicircle: $ P = \pi r + 2r $ or $ P = r(\pi + 2) $

This includes the curved arc ($ \pi r $) plus the diameter ($ 2r $)

Quarter Circle:

Area of Quarter Circle: $ A = \frac{1}{4}\pi r^2 $ or $ A = \frac{\pi r^2}{4} $

Perimeter of Quarter Circle: $ P = \frac{1}{2}\pi r + 2r $ or $ P = \frac{\pi r}{2} + 2r $

This includes the curved arc ($ \frac{\pi r}{2} $) plus the two radii ($ 2r $)

Examples:

Example 1: Find the area and perimeter of a semicircle with radius 10 cm. (Use $ \pi = 3.14 $)

Area: $ A = \frac{1}{2}\pi r^2 = \frac{1}{2} \times 3.14 \times 10^2 = \frac{1}{2} \times 314 = 157 $ cm²

Perimeter: $ P = \pi r + 2r = (3.14 \times 10) + (2 \times 10) = 31.4 + 20 = 51.4 $ cm

Example 2: Find the area of a quarter circle with radius 14 cm. (Use $ \pi = \frac{22}{7} $)

$ A = \frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times \frac{22}{7} \times 196 $

$ A = \frac{1}{4} \times 616 = 154 $ cm²

Example 3: A quarter circle has radius 6 m. Find its perimeter. (Use $ \pi = 3.14 $)

$ P = \frac{\pi r}{2} + 2r = \frac{3.14 \times 6}{2} + (2 \times 6) = 9.42 + 12 = 21.42 $ m

6. Circles: Word Problems

Examples:

Example 1: A circular pizza has a diameter of 16 inches. How many square inches of pizza are there? (Use $ \pi = 3.14 $)

Radius = $ \frac{16}{2} = 8 $ inches

Area = $ \pi r^2 = 3.14 \times 8^2 = 3.14 \times 64 = 200.96 $ square inches

Example 2: A circular running track has a radius of 50 meters. How far does a runner travel going around it once? (Use $ \pi = 3.14 $)

Looking for: Circumference

$ C = 2\pi r = 2 \times 3.14 \times 50 = 314 $ meters

Example 3: A circular pond has a circumference of 88 meters. Find its radius. (Use $ \pi = \frac{22}{7} $)

$ C = 2\pi r $

$ 88 = 2 \times \frac{22}{7} \times r $

$ 88 = \frac{44r}{7} $

$ r = \frac{88 \times 7}{44} = 14 $ meters

Example 4: A semicircular window has a diameter of 4 feet. How much glass is needed to fill it? (Use $ \pi = 3.14 $)

Radius = $ \frac{4}{2} = 2 $ feet

Area = $ \frac{1}{2}\pi r^2 = \frac{1}{2} \times 3.14 \times 2^2 = \frac{1}{2} \times 12.56 = 6.28 $ ft²

7. Area Between Two Shapes

Concept: The area between two shapes is found by subtracting the area of the smaller (inner) shape from the area of the larger (outer) shape.

Formula:

Area Between Shapes = Area of Outer Shape - Area of Inner Shape

Common Scenarios:

Ring/Annulus: Area between two concentric circles
Frame: Area between two rectangles
Path/Border: Area of walkway around a shape
Shaded region problems

Examples:

Example 1 (Ring/Annulus): Find the area of a ring formed by two concentric circles with outer radius 10 cm and inner radius 6 cm. (Use $ \pi = 3.14 $)

Outer circle area: $ A_1 = \pi r^2 = 3.14 \times 10^2 = 314 $ cm²

Inner circle area: $ A_2 = \pi r^2 = 3.14 \times 6^2 = 113.04 $ cm²

Area of ring: $ 314 - 113.04 = 200.96 $ cm²

Example 2 (Frame): A picture frame has outer dimensions 20 cm by 16 cm and inner dimensions 14 cm by 10 cm. Find the area of the frame.

Outer rectangle area: $ 20 \times 16 = 320 $ cm²

Inner rectangle area: $ 14 \times 10 = 140 $ cm²

Frame area: $ 320 - 140 = 180 $ cm²

Example 3 (Path): A rectangular garden is 30 m by 20 m with a 2-meter wide path around it. Find the area of the path.

Garden area: $ 30 \times 20 = 600 $ m²

Garden + Path dimensions: $ (30 + 4) \times (20 + 4) = 34 \times 24 $

Total area: $ 34 \times 24 = 816 $ m²

Path area: $ 816 - 600 = 216 $ m²

Example 4 (Shaded Region): A circle with radius 5 cm is inscribed in a square. Find the area of the shaded region (square minus circle). (Use $ \pi = 3.14 $)

Square side: $ 2r = 10 $ cm

Square area: $ 10^2 = 100 $ cm²

Circle area: $ \pi r^2 = 3.14 \times 5^2 = 78.5 $ cm²

Shaded area: $ 100 - 78.5 = 21.5 $ cm²

8. Perimeter and Area: Changes in Scale

Key Concept: When a figure is scaled by a factor $ k $, perimeter and area change in predictable ways.

Scale Factor Effects:

If scale factor = $ k $:

Linear dimensions (length, width, height) multiply by $ k $
Perimeter multiplies by $ k $
Area multiplies by $ k^2 $
Volume multiplies by $ k^3 $

Measure	Scale Effect	Formula
Length	Multiply by k	$ \text{New length} = k \times \text{Old length} $
Perimeter	Multiply by k	$ \text{New perimeter} = k \times \text{Old perimeter} $
Area	Multiply by k²	$ \text{New area} = k^2 \times \text{Old area} $