Area of Rectangles & Decomposed Shapes - Fourth Grade Mathematics
When farmers calculate the planting capacity of a wheat field, when tile-setters determine how many ceramic tiles are needed to cover a kitchen floor, or when landscape designers lay fresh sod across a park, they are calculating area. Area is the measure of the total two-dimensional surface enclosed inside the boundary of a shape, measured in square units. In fourth grade, you transition from physically counting grid squares to utilizing the algebraic formula Area = Length x Width, discovering the inverse relationship between area and perimeter, and mastering the decomposition of complex composite figures into non-overlapping rectangles.
The Area Formula for Rectangles
The area of any rectangle is determined by multiplying its length by its width.
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| THE AREA OF A RECTANGLE FORMULA |
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| Formula: Area = Length x Width (A = L x W) |
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| Example: A rectangle measuring 7 meters by 4 meters |
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| <----------------- Length = 7 m -----------------> |
| +-----+-----+-----+-----+-----+-----+-----+ |
| | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | Row 1 (7 square meters) |
| Width +-----+-----+-----+-----+-----+-----+-----+ |
| = 4 m | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | Row 2 (7 square meters) |
| +-----+-----+-----+-----+-----+-----+-----+ |
| | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | Row 3 (7 square meters) |
| +-----+-----+-----+-----+-----+-----+-----+ |
| | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | 1m2 | Row 4 (7 square meters) |
| +-----+-----+-----+-----+-----+-----+-----+ |
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| Notice: There are 4 rows of 7 square meters each! |
| Area = 4 x 7 = 28 square meters (28 m^2) |
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Why Units Must Be Squared
When you calculate perimeter, you add lengths together: 7 m + 4 m = 11 m (linear meters). But when you calculate area, you multiply meters by meters: meters x meters = square meters (m^2). A square meter is a two-dimensional physical tile measuring 1 meter on each side. Always label area answers with square units: square inches (sq. in.), square feet (sq. ft.), or square meters (sq. m.).
Finding a Missing Dimension from Known Area
If you know the total area of a rectangle and one of its dimensions, you use the inverse operation of division to find the missing side.
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| FINDING A MISSING SIDE FROM AREA |
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| Problem: A rectangular rug has an Area of 72 square feet and a Width of 8 feet. |
| What is the Length (L)? |
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| Formula : Area = Length x Width |
| Equation: 72 = L x 8 |
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| Solve using Division: |
| L = Area / Width |
| L = 72 / 8 = 9 feet! |
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| Check: 9 ft x 8 ft = 72 sq ft. Correct! |
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Decomposing Composite Figures to Find Total Area
To find the area of an irregular or composite rectilinear shape (such as an L-shaped floor), you decompose the shape into smaller, non-overlapping rectangles, compute the area of each room, and add them together.
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| DECOMPOSING AN L-SHAPE TO FIND AREA |
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| <------------- 10 m -------------> |
| +----------------------------------------+ |
| | | 3 m |
| 8 m | ROOM A (Top Rectangle) | |
| +-----------------------+----------------+ |
| | | 6 m |
| | ROOM B | |
| | (Bottom Rectangle) | 5 m |
| +-----------------------+ |
| 6 m |
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| METHOD 1 (Horizontal Cut): |
| Room A (Top) : Length = 10 m, Width = 3 m. |
| Area A = 10 x 3 = 30 square meters. |
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| Room B (Bottom): Height = 8 - 3 = 5 m, Width = 6 m. |
| Area B = 6 x 5 = 30 square meters. |
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| Total Area = Area A + Area B = 30 + 30 = 60 square meters! |
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| METHOD 2 (Subtraction from Outer Bounding Box): |
| Full outer box = 10 x 8 = 80 sq m. |
| Cut-out notch = 4 x 5 = 20 sq m (where 10 - 6 = 4 m). |
| Total Area = 80 - 20 = 60 square meters! |
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| Both methods produce the exact same total area of 60 square meters! |
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Chapter Practice Exercises
Exercise 1: Compute the area of a rectangular patio that has a length of 14 feet and a width of 8 feet. Include proper square units.
Exercise 2: A rectangular banner has an area of 108 square feet. If the width of the banner is 9 feet, what is its length?
Exercise 3: A square tile has side lengths of 12 inches. Find both the perimeter and the area of the tile, comparing the numerical values and units.
Exercise 4: An L-shaped room is decomposed into two rectangles: Rectangle 1 measures 8 meters by 5 meters, and Rectangle 2 measures 4 meters by 3 meters. What is the total area of the room?
Exercise 5: A student looks at a rectangle with an area of 36 square feet and claims its perimeter must be 26 feet because 9 x 4 = 36 and 2 x (9 + 4) = 26. Is 26 feet the only possible perimeter for a rectangle with an area of 36 square feet? Explain and provide another valid example.
Solutions and Step-by-Step Explanations
Solution 1: Using the formula Area = Length x Width: Area = 14 ft x 8 ft = 112 square feet (sq. ft.).
Solution 2: We use the inverse relationship Length = Area / Width: Length = 108 / 9 = 12 feet. The banner is 12 feet long.
Solution 3: For a square of side 12 inches: Perimeter = 4 x 12 = 48 inches (linear distance). Area = 12 x 12 = 144 square inches (two-dimensional surface). Notice that perimeter measures length in inches, while area measures enclosed surface in square inches; their units and numerical values represent completely different physical concepts.
Solution 4: Find the area of each decomposed room: Area 1 = 8 m x 5 m = 40 square meters. Area 2 = 4 m x 3 m = 12 square meters. Adding the areas: 40 + 12 = 52 square meters. The total area of the room is 52 square meters.
Solution 5: No, 26 feet is not the only possible perimeter. Many different rectangles can share the exact same area while having completely different perimeters. For an area of 36 square feet, the dimensions could be: 6 ft by 6 ft (Area = 36 sq ft, Perimeter = 4 x 6 = 24 feet). 12 ft by 3 ft (Area = 36 sq ft, Perimeter = 2 x (12 + 3) = 30 feet). 18 ft by 2 ft (Area = 36 sq ft, Perimeter = 2 x (18 + 2) = 40 feet). 36 ft by 1 ft (Area = 36 sq ft, Perimeter = 2 x (36 + 1) = 74 feet). Rectangles with the same area can have widely varying perimeters.