Learn on PengiThe Art of Problem Solving: Prealgebra (AMC 8)Chapter 11: Perimeter and Area

Lesson 2: Area

In this Grade 4 lesson from The Art of Problem Solving: Prealgebra (AMC 8), students learn how to calculate area using the formulas for rectangles, right triangles, and general triangles with a base and altitude. The lesson introduces key concepts such as unit squares, the rectangle area formula (l × w), and the triangle area formula (bh/2), including how to identify and draw altitudes. Students also develop problem-solving strategies for finding areas of complex shapes by decomposing them into rectangles and triangles or using subtraction of known areas.

Section 1

Area

Property

The number of squares that fit inside a perimeter is called the area of the region enclosed. A square unit is a square that measures 1 unit on each side. Perimeter measures the distance around the outside of a region, while Area measures the amount of space enclosed inside the region.

Examples

A rectangular piece of paper is 8 inches wide and 11 inches long. Its area is calculated by multiplying the dimensions: $8 \times 11 = 88$ square inches.
An L-shaped patio is made of two rectangles. One is $4 \text{ m} \times 3 \text{ m}$ and the other is $5 \text{ m} \times 2 \text{ m}$ . The total area is $(4 \times 3) + (5 \times 2) = 12 + 10 = 22$ square meters.
A kitchen floor is 12 feet long and 10 feet wide. The area is $12 \times 10 = 120$ square feet.

Explanation

Area tells you how much surface a shape covers. Think of it as the amount of carpet needed for a floor or paint for a wall. The more square units that fit inside a shape, the larger its area.

Section 2

Area of a Rectangle

Property

For a rectangle with length $L$ and width $W$ , the area, $A$ , is given by the formula:

A = L \cdot W

This formula is used when solving problems involving rectangular shapes where the dimensions are related to each other.

Examples

A rectangular garden has an area of 176 square feet. Its length is 5 feet more than its width. Let width be $w$ . The equation is $w(w+5)=176$ , or $w^2+5w-176=0$ . Solving gives $w=11$ . The width is 11 ft and the length is 16 ft.
The area of a rectangular patio is 250 square meters. The length is twice the width. Let width be $w$ . The equation is $w(2w)=250$ , or $2w^2=250$ . This gives $w^2=125$ , so $w = \sqrt{125} \approx 11.2$ . The width is about 11.2 m and the length is about 22.4 m.
A rectangular screen has an area of 90 square inches. Its width is 1 inch less than half its length. Let length be $L$ . The equation is $L(0.5L-1)=90$ , so $0.5L^2-L-90=0$ . This gives $L^2-2L-180=0$ . Solving gives $L \approx 14.4$ . The length is about 14.4 in and the width is about 6.2 in.

Explanation

This formula helps find unknown dimensions of a rectangle. When length is expressed in terms of width, substituting into the area formula creates a quadratic equation. Solving it reveals the exact measurements for the length and width.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Area

Property

Examples

A rectangular piece of paper is 8 inches wide and 11 inches long. Its area is calculated by multiplying the dimensions: $8 \times 11 = 88$ square inches.
An L-shaped patio is made of two rectangles. One is $4 \text{ m} \times 3 \text{ m}$ and the other is $5 \text{ m} \times 2 \text{ m}$ . The total area is $(4 \times 3) + (5 \times 2) = 12 + 10 = 22$ square meters.
A kitchen floor is 12 feet long and 10 feet wide. The area is $12 \times 10 = 120$ square feet.

Explanation

Area tells you how much surface a shape covers. Think of it as the amount of carpet needed for a floor or paint for a wall. The more square units that fit inside a shape, the larger its area.

Section 2

Area of a Rectangle

Property

For a rectangle with length $L$ and width $W$ , the area, $A$ , is given by the formula:

A = L \cdot W

This formula is used when solving problems involving rectangular shapes where the dimensions are related to each other.

Examples

A rectangular garden has an area of 176 square feet. Its length is 5 feet more than its width. Let width be $w$ . The equation is $w(w+5)=176$ , or $w^2+5w-176=0$ . Solving gives $w=11$ . The width is 11 ft and the length is 16 ft.
The area of a rectangular patio is 250 square meters. The length is twice the width. Let width be $w$ . The equation is $w(2w)=250$ , or $2w^2=250$ . This gives $w^2=125$ , so $w = \sqrt{125} \approx 11.2$ . The width is about 11.2 m and the length is about 22.4 m.
A rectangular screen has an area of 90 square inches. Its width is 1 inch less than half its length. Let length be $L$ . The equation is $L(0.5L-1)=90$ , so $0.5L^2-L-90=0$ . This gives $L^2-2L-180=0$ . Solving gives $L \approx 14.4$ . The length is about 14.4 in and the width is about 6.2 in.

Explanation

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Area

Property

Examples

A rectangular piece of paper is 8 inches wide and 11 inches long. Its area is calculated by multiplying the dimensions: $8 \times 11 = 88$ square inches.
An L-shaped patio is made of two rectangles. One is $4 \text{ m} \times 3 \text{ m}$ and the other is $5 \text{ m} \times 2 \text{ m}$ . The total area is $(4 \times 3) + (5 \times 2) = 12 + 10 = 22$ square meters.
A kitchen floor is 12 feet long and 10 feet wide. The area is $12 \times 10 = 120$ square feet.

Explanation

Area tells you how much surface a shape covers. Think of it as the amount of carpet needed for a floor or paint for a wall. The more square units that fit inside a shape, the larger its area.

Section 2

Area of a Rectangle

Property

For a rectangle with length $L$ and width $W$ , the area, $A$ , is given by the formula:

A = L \cdot W

This formula is used when solving problems involving rectangular shapes where the dimensions are related to each other.

Examples

A rectangular garden has an area of 176 square feet. Its length is 5 feet more than its width. Let width be $w$ . The equation is $w(w+5)=176$ , or $w^2+5w-176=0$ . Solving gives $w=11$ . The width is 11 ft and the length is 16 ft.
The area of a rectangular patio is 250 square meters. The length is twice the width. Let width be $w$ . The equation is $w(2w)=250$ , or $2w^2=250$ . This gives $w^2=125$ , so $w = \sqrt{125} \approx 11.2$ . The width is about 11.2 m and the length is about 22.4 m.
A rectangular screen has an area of 90 square inches. Its width is 1 inch less than half its length. Let length be $L$ . The equation is $L(0.5L-1)=90$ , so $0.5L^2-L-90=0$ . This gives $L^2-2L-180=0$ . Solving gives $L \approx 14.4$ . The length is about 14.4 in and the width is about 6.2 in.