Area and Perimeter

Measure the space inside and the distance around

You use area and perimeter more often than you might realize. When you measure a room to buy carpet, you are calculating area. When you figure out how much fencing you need for your backyard, that is perimeter. When you order just the right amount of pizza for a party, you are thinking about area (even if you do not consciously run the numbers). These are not abstract mathematical concepts locked away in textbooks - they are tools you already use to navigate everyday decisions. Now we are going to sharpen those tools and give you precise methods for any shape you encounter.

Core Concepts

Perimeter: Walking Around the Edge

Perimeter is the total distance around the outside of a shape. Picture yourself walking along the boundary of a park. When you return to your starting point, the total distance you walked is the perimeter.

For any polygon (a shape with straight sides), finding the perimeter is straightforward: add up the lengths of all the sides.

Rectangle: Since opposite sides are equal, you have two lengths and two widths: $$P = 2l + 2w = 2(l + w)$$

Square: All four sides are the same length $s$: $$P = 4s$$

Any polygon: Just add all the sides together: $$P = s_1 + s_2 + s_3 + \cdots$$

Circumference: Perimeter’s Circular Cousin

When the shape is a circle, we give the perimeter a special name: circumference. The relationship between a circle’s circumference and its diameter is one of the most beautiful constants in mathematics - pi ($\pi$), approximately $3.14159$.

$$C = \pi d = 2\pi r$$

where $d$ is the diameter and $r$ is the radius. These formulas say the same thing since $d = 2r$.

Area: The Space Inside

While perimeter measures distance around, area measures the space enclosed within a shape. If perimeter is the fence, area is the yard inside. We measure area in square units - square feet, square meters, square inches, and so on - because we are essentially counting how many unit squares fit inside the shape.

Area of Rectangles

The simplest area formula is for a rectangle: $$A = l \times w$$

Multiply the length by the width, and you have the area. This makes intuitive sense: if you have a rectangle that is 4 units long and 3 units wide, you could lay out a grid of $4 \times 3 = 12$ unit squares inside it.

For a square where all sides equal $s$: $$A = s^2$$

Area of Triangles

A triangle’s area is half of a related rectangle’s area. Here is why: take any triangle, make a copy of it, rotate the copy, and fit the two together. You get a parallelogram (which we will see has the same area as a rectangle with the same base and height). Since your triangle is half of that parallelogram:

$$A = \frac{1}{2} \times b \times h$$

Important: The height $h$ must be perpendicular (at a right angle) to the base $b$. The height is not necessarily one of the sides of the triangle - it is the straight-line distance from the base to the opposite vertex.

Area of Parallelograms

A parallelogram is a four-sided figure where opposite sides are parallel. If you slice off a triangle from one end of a parallelogram and move it to the other end, you get a rectangle. This shows that:

$$A = b \times h$$

Again, the height $h$ must be perpendicular to the base. It is not the slanted side, but the perpendicular distance between the two parallel bases.

Area of Trapezoids

A trapezoid has exactly one pair of parallel sides, called the bases ($b_1$ and $b_2$). To find its area, we average the two bases and multiply by the height:

$$A = \frac{1}{2}(b_1 + b_2) \times h$$

Why does this work? Think of it as the average of two rectangles - one with the smaller base and one with the larger base. The trapezoid’s area falls exactly in between.

Another way to see it: you can divide a trapezoid into two triangles, each with height $h$. One has base $b_1$, the other has base $b_2$. Adding their areas: $\frac{1}{2}b_1h + \frac{1}{2}b_2h = \frac{1}{2}(b_1 + b_2)h$.

Area of Regular Polygons

A regular polygon has all sides equal and all angles equal. Think of regular hexagons (like honeycomb cells), regular pentagons (like home plate in baseball, roughly), or regular octagons (like stop signs).

For regular polygons, we use the apothem - the perpendicular distance from the center of the polygon to the middle of any side. If you know the apothem $a$ and the perimeter $P$:

$$A = \frac{1}{2} \times a \times P$$

Why does this work? Imagine drawing lines from the center to each vertex, dividing the polygon into identical triangles. Each triangle has a base (one side of the polygon) and a height (the apothem). The total area is the sum of all these triangles: $$A = n \times \frac{1}{2} \times s \times a = \frac{1}{2} \times a \times (n \times s) = \frac{1}{2} \times a \times P$$

where $n$ is the number of sides and $s$ is the side length.

Area of Circles

The area of a circle with radius $r$ is:

$$A = \pi r^2$$

This formula says: take the radius, square it, and multiply by $\pi$. If you double the radius, the area does not double - it quadruples (because $2^2 = 4$).

Where does $\pi r^2$ come from? One beautiful way to see it: cut a circle into many thin wedges (like pizza slices) and rearrange them into something that looks like a parallelogram. The “height” of this parallelogram is the radius $r$, and the “base” is half the circumference, or $\pi r$. Area = base times height = $\pi r \times r = \pi r^2$.

Area of Sectors

A sector is a “pie slice” of a circle - the region between two radii and an arc. If the central angle of the sector is $\theta$ (in degrees), then the sector is $\frac{\theta}{360}$ of the full circle. So:

$$A_{\text{sector}} = \frac{\theta}{360} \times \pi r^2$$

For example, a $90°$ sector is one-quarter of the circle (since $\frac{90}{360} = \frac{1}{4}$), so its area is $\frac{1}{4}\pi r^2$.

If you are working in radians (where a full circle is $2\pi$ radians instead of $360°$):

$$A_{\text{sector}} = \frac{1}{2}r^2\theta$$

Composite Figures

Real-world shapes are rarely perfect rectangles or circles. A floor plan might be L-shaped. A garden might combine rectangular and semicircular sections. These are composite figures - shapes made by combining (or subtracting) simpler shapes.

Strategy for composite figures:

  1. Identify the simpler shapes that make up the figure
  2. Calculate the area of each part
  3. Add (if parts combine) or subtract (if there are holes or cutouts)

The same approach works for perimeter, but be careful: when shapes combine, some edges become interior and do not count toward the outer perimeter.

The Relationship Between Perimeter and Area

A common misconception is that shapes with the same perimeter have the same area, or vice versa. This is not true.

Consider two rectangles:

  • Rectangle A: 10 by 10 (a square). Perimeter = 40, Area = 100
  • Rectangle B: 19 by 1. Perimeter = 40, Area = 19

Same perimeter, wildly different areas! In fact, for a fixed perimeter, a square maximizes the area. For a fixed area, a square minimizes the perimeter.

Among all shapes with a fixed perimeter, a circle encloses the most area. This is why bubbles are round - nature found the shape that maximizes volume for a given surface area.

Notation and Terminology

Term Meaning Example
Perimeter Distance around a shape $P$
Area Space inside a shape $A$
Circumference Perimeter of a circle $C = 2\pi r$
Base A side used as reference for height $b$
Height Perpendicular distance to the base $h$
Apothem Distance from center to middle of side (regular polygon) $a$
Sector “Pie slice” region of a circle
Central angle Angle at center of circle forming a sector $\theta$
Composite figure Shape made of multiple simpler shapes
Pi ($\pi$) Ratio of circumference to diameter $\approx 3.14159$
Square units Units for measuring area sq ft, $\text{m}^2$

Examples

Example 1: Area of a Triangle

Find the area of a triangle with a base of 12 cm and a height of 7 cm.

Solution:

The area of a triangle is half the base times the height: $$A = \frac{1}{2} \times b \times h$$

Substitute the given values: $$A = \frac{1}{2} \times 12 \times 7$$ $$A = \frac{1}{2} \times 84$$ $$A = 42 \text{ square centimeters}$$

The triangle has an area of 42 cm².

Why this makes sense: A rectangle with the same base and height would be $12 \times 7 = 84$ cm². The triangle takes up exactly half of that rectangle, so 42 cm² is correct.

Example 2: Area of a Parallelogram

A parallelogram has a base of 15 inches and a height of 8 inches. The slanted side measures 10 inches. Find the area.

Solution:

For a parallelogram, the formula is simply base times height: $$A = b \times h$$

Notice that we do not use the slanted side length (10 inches) at all. The area depends only on the base and the perpendicular height: $$A = 15 \times 8 = 120 \text{ square inches}$$

The parallelogram has an area of 120 in².

Common mistake to avoid: Do not multiply the base by the slanted side. That would give $15 \times 10 = 150$ in², which is too large. The height must be perpendicular to the base.

Example 3: Area of a Regular Hexagon Using the Apothem

A regular hexagon has sides of length 6 cm and an apothem of approximately 5.2 cm. Find its area.

Solution:

For a regular polygon, we use the formula: $$A = \frac{1}{2} \times a \times P$$

where $a$ is the apothem and $P$ is the perimeter.

Step 1: Find the perimeter. A hexagon has 6 sides, each of length 6 cm: $$P = 6 \times 6 = 36 \text{ cm}$$

Step 2: Apply the formula: $$A = \frac{1}{2} \times 5.2 \times 36$$ $$A = \frac{1}{2} \times 187.2$$ $$A = 93.6 \text{ square centimeters}$$

The regular hexagon has an area of approximately 93.6 cm².

Why the apothem formula works: Imagine drawing lines from the center to each vertex, creating 6 identical triangles. Each triangle has base 6 cm and height 5.2 cm (the apothem). Each triangle’s area is $\frac{1}{2}(6)(5.2) = 15.6$ cm². Six triangles give $6 \times 15.6 = 93.6$ cm².

Example 4: Area of a Sector

A pizza has a diameter of 16 inches. You eat a slice where the crust edge (arc) corresponds to a central angle of 45°. What is the area of your slice?

Solution:

First, identify what we know:

  • Diameter = 16 inches, so radius $r = 8$ inches
  • Central angle $\theta = 45°$

The sector area formula is: $$A_{\text{sector}} = \frac{\theta}{360} \times \pi r^2$$

Step 1: Calculate what fraction of the pizza your slice represents: $$\frac{45}{360} = \frac{1}{8}$$

Your slice is one-eighth of the whole pizza.

Step 2: Calculate the full pizza’s area: $$A_{\text{pizza}} = \pi r^2 = \pi (8)^2 = 64\pi \approx 201.1 \text{ square inches}$$

Step 3: Find the sector area: $$A_{\text{sector}} = \frac{1}{8} \times 64\pi = 8\pi \approx 25.1 \text{ square inches}$$

Your pizza slice has an area of approximately 25.1 square inches.

Quick check: $45° \times 8 = 360°$, so 8 slices of this size would make a whole pizza. The whole pizza is about 201 in², and $201 \div 8 \approx 25$. That matches.

Example 5: Area of a Composite Figure

A room has the following shape: a rectangle that is 20 feet long and 15 feet wide, with a semicircular bay window extending outward from one of the 15-foot walls. The bay window has a diameter of 8 feet. Find the total floor area of the room.

Solution:

This is a composite figure: a rectangle plus a semicircle.

Step 1: Sketch the situation (mentally or on paper). We have:

  • A 20 ft by 15 ft rectangle
  • A semicircle with diameter 8 ft attached to the 15 ft wall

Step 2: Calculate the rectangular area: $$A_{\text{rectangle}} = l \times w = 20 \times 15 = 300 \text{ square feet}$$

Step 3: Calculate the semicircular area.

The diameter is 8 ft, so the radius is $r = 4$ ft.

A full circle’s area would be: $$A_{\text{circle}} = \pi r^2 = \pi (4)^2 = 16\pi$$

A semicircle is half of that: $$A_{\text{semicircle}} = \frac{1}{2} \times 16\pi = 8\pi \approx 25.1 \text{ square feet}$$

Step 4: Add the areas together: $$A_{\text{total}} = 300 + 8\pi \approx 300 + 25.1 = 325.1 \text{ square feet}$$

The total floor area is approximately 325.1 square feet (or exactly $300 + 8\pi$ square feet).

For practical purposes: If you are buying flooring, round up to 330 square feet to account for waste and cuts.

Example 6: Composite Figure with Subtraction

A rectangular metal sheet measures 24 inches by 18 inches. Four identical circles with radius 3 inches are cut out (one near each corner). What is the remaining area of the metal sheet?

Solution:

Here we need to subtract the cut-out areas from the total.

Step 1: Calculate the original rectangular area: $$A_{\text{rectangle}} = 24 \times 18 = 432 \text{ square inches}$$

Step 2: Calculate the area of one circular cutout: $$A_{\text{circle}} = \pi r^2 = \pi (3)^2 = 9\pi \approx 28.3 \text{ square inches}$$

Step 3: Calculate the total area of all four cutouts: $$A_{\text{4 circles}} = 4 \times 9\pi = 36\pi \approx 113.1 \text{ square inches}$$

Step 4: Subtract to find the remaining area: $$A_{\text{remaining}} = 432 - 36\pi \approx 432 - 113.1 = 318.9 \text{ square inches}$$

The remaining metal area is approximately 318.9 square inches (or exactly $432 - 36\pi$ square inches).

Key Properties and Rules

Perimeter Formulas

Rectangle: $$P = 2l + 2w = 2(l + w)$$

Square: $$P = 4s$$

Triangle: $$P = a + b + c$$

Regular polygon with $n$ sides of length $s$: $$P = ns$$

Circle (circumference): $$C = \pi d = 2\pi r$$

Area Formulas

Rectangle: $$A = l \times w$$

Square: $$A = s^2$$

Triangle: $$A = \frac{1}{2}bh$$

Parallelogram: $$A = bh$$

Trapezoid: $$A = \frac{1}{2}(b_1 + b_2)h$$

Regular polygon: $$A = \frac{1}{2}aP$$ where $a$ = apothem, $P$ = perimeter

Circle: $$A = \pi r^2$$

Sector (degrees): $$A = \frac{\theta}{360} \times \pi r^2$$

Sector (radians): $$A = \frac{1}{2}r^2\theta$$

Critical Reminders

  1. Height is always perpendicular. For triangles, parallelograms, and trapezoids, the height forms a right angle with the base. It is not necessarily a side of the shape.

  2. Use the right measurement. Circles use radius (not diameter) in the area formula. Confusing these gives you an answer that is off by a factor of 4.

  3. Watch your units. If lengths are in feet, area is in square feet. If you mix inches and feet, your answer will be wrong.

  4. Composite figures require strategy. Break them into pieces, calculate each piece, then add or subtract appropriately.

  5. Same perimeter does not mean same area. A circle maximizes area for a given perimeter. Among rectangles, a square maximizes area for a given perimeter.

Real-World Applications

Flooring and Carpeting

When replacing the flooring in a room, you need to calculate the floor area to know how much material to buy. If the room is not a simple rectangle - if it has closets, bay windows, or L-shaped layouts - you need to break it into composite shapes. Always buy a bit extra (typically 10%) to account for waste from cuts.

Painting Walls

Paint coverage is measured in square feet per gallon. To figure out how much paint you need, calculate the wall area (height times width for each wall), subtract for windows and doors, and divide by the coverage rate. Knowing perimeter also helps - if you are installing crown molding or baseboards, you need to know the perimeter of the room.

Landscaping

Landscapers constantly calculate area for sod, mulch, and ground cover. Sod is sold by the square foot. Mulch is sold by the cubic yard but is applied at a certain depth, so you start by calculating the area to cover. Irregular garden beds might be approximated as combinations of rectangles, triangles, and circles.

Fabric for Sewing

Patterns list how much fabric you need, but understanding area helps you verify those numbers or work with irregularly shaped pieces. A quilt with many small pieces requires careful area calculations to minimize waste.

Construction and Architecture

Calculating roof area (often involving triangles and trapezoids) determines how many shingles you need. Foundation area affects concrete requirements. Window area affects energy efficiency calculations. Architects and contractors work with area and perimeter constantly.

Pizza Pricing

Here is a fun one: is a large pizza a better deal than two mediums? If a large has a 16-inch diameter and a medium has a 12-inch diameter:

  • Large area: $\pi(8)^2 = 64\pi \approx 201$ square inches
  • Two mediums: $2 \times \pi(6)^2 = 72\pi \approx 226$ square inches

Two mediums actually give you more pizza! But restaurants know this, so they often price accordingly.

Self-Test Problems

Problem 1: A rectangular garden is 25 feet long and 18 feet wide. You want to put a fence around it and also plant grass inside. How much fencing do you need (perimeter)? How much sod do you need (area)?

Show Answer

Perimeter (fencing): $$P = 2l + 2w = 2(25) + 2(18) = 50 + 36 = 86 \text{ feet}$$

You need 86 feet of fencing.

Area (sod): $$A = l \times w = 25 \times 18 = 450 \text{ square feet}$$

You need 450 square feet of sod.

Problem 2: A triangular sail has a base of 14 feet and a height of 22 feet. What is the area of the sail?

Show Answer

$$A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 14 \times 22 = \frac{1}{2} \times 308 = 154 \text{ square feet}$$

The sail has an area of 154 square feet.

Problem 3: A trapezoid-shaped plot of land has parallel sides of 80 meters and 120 meters, with a perpendicular distance of 50 meters between them. What is the area of the plot?

Show Answer

$$A = \frac{1}{2}(b_1 + b_2) \times h = \frac{1}{2}(80 + 120) \times 50$$ $$A = \frac{1}{2}(200) \times 50 = 100 \times 50 = 5000 \text{ square meters}$$

The plot has an area of 5000 square meters (or 0.5 hectares).

Problem 4: A regular octagon (8-sided figure) has sides of length 5 inches and an apothem of approximately 6.04 inches. Find its area.

Show Answer

Step 1: Find the perimeter. $$P = 8 \times 5 = 40 \text{ inches}$$

Step 2: Apply the regular polygon area formula. $$A = \frac{1}{2} \times a \times P = \frac{1}{2} \times 6.04 \times 40$$ $$A = \frac{1}{2} \times 241.6 = 120.8 \text{ square inches}$$

The regular octagon has an area of approximately 120.8 square inches.

Problem 5: A circular fountain has a radius of 6 meters. A walkway in the shape of a sector with a central angle of 60° extends from the fountain. What is the area of this walkway sector?

Show Answer

$$A_{\text{sector}} = \frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \pi (6)^2$$ $$A_{\text{sector}} = \frac{1}{6} \times 36\pi = 6\pi \approx 18.85 \text{ square meters}$$

The sector walkway has an area of approximately 18.85 square meters (or exactly $6\pi$ square meters).

Problem 6: An L-shaped room consists of two rectangles: one that is 12 feet by 10 feet, and another that is 8 feet by 6 feet. If the two rectangles share a common wall (the 8-foot side of the smaller rectangle meets the 10-foot side of the larger rectangle), what is the total floor area?

Show Answer

For an L-shaped room, we simply add the areas of the two rectangular sections:

Larger rectangle: $$A_1 = 12 \times 10 = 120 \text{ square feet}$$

Smaller rectangle: $$A_2 = 8 \times 6 = 48 \text{ square feet}$$

Total area: $$A_{\text{total}} = 120 + 48 = 168 \text{ square feet}$$

The L-shaped room has an area of 168 square feet.

Problem 7: A circular pool has a diameter of 24 feet. A concrete deck 3 feet wide surrounds the pool. What is the area of just the concrete deck (not including the pool)?

Show Answer

This is a composite figure problem involving subtraction.

Pool radius: $r_{\text{pool}} = 12$ feet

Outer radius (pool + deck): $r_{\text{outer}} = 12 + 3 = 15$ feet

Area of larger circle (pool + deck): $$A_{\text{outer}} = \pi (15)^2 = 225\pi$$

Area of pool: $$A_{\text{pool}} = \pi (12)^2 = 144\pi$$

Area of deck alone: $$A_{\text{deck}} = 225\pi - 144\pi = 81\pi \approx 254.5 \text{ square feet}$$

The concrete deck has an area of approximately 254.5 square feet.

Summary

Perimeter measures the distance around a shape. For polygons, add all the sides. For circles, the perimeter is called circumference: $C = 2\pi r = \pi d$.

Area measures the space inside a shape. Key formulas:

  • Rectangle: $A = lw$
  • Triangle: $A = \frac{1}{2}bh$ (height must be perpendicular to base)
  • Parallelogram: $A = bh$ (height must be perpendicular to base)
  • Trapezoid: $A = \frac{1}{2}(b_1 + b_2)h$
  • Regular polygon: $A = \frac{1}{2}aP$ (where $a$ is the apothem)
  • Circle: $A = \pi r^2$
  • Sector: $A = \frac{\theta}{360} \times \pi r^2$ (for angle in degrees)

For composite figures, break the shape into simpler parts, calculate each area, and add or subtract as needed.

Remember that perimeter and area are independent. Shapes with the same perimeter can have very different areas, and vice versa. Among all shapes with a given perimeter, a circle encloses the maximum area.

These formulas are not just academic exercises - they are practical tools. Whether you are buying paint, laying carpet, planning a garden, or just figuring out which pizza is the best deal, area and perimeter calculations help you make informed decisions. The more you practice, the more natural these calculations become, until you find yourself estimating areas without even thinking about it.