Rectangle Dimensions and Area: Solving for Length, Width, and Perimeter with Examples

Understanding how to calculate the dimensions and area of a rectangle is fundamental in geometry. This guide covers various examples, including the relationship between length and width, perimeter calculation, and area determination. Whether you're a student or a professional, these examples will provide a solid foundation for solving similar problems efficiently.

Example 1: Length is 3 Times the Width

Given:

The length of a rectangle is 3 times its width. The perimeter of the rectangle is 63 inches.

Steps:

Define the width as w. The length, L, is 3w. Use the perimeter formula: P 2L 2w. Substitute the values: 63 2(3w) 2w. Simplify: 63 6w 2w or 63 8w. Solve for w: w frac{63}{8} 7.875 inches. Calculate the length: l 3w 3 times 7.875 23.625 inches. Calculate the area: A l times w 23.625 times 7.875 or A approximately 186.328125 square inches.

Conclusion: The area of the rectangle is approximately 186.33 square inches.

Example 2: Length is 4 Times the Width

Given:

The length of a rectangle is 4 times its width. The perimeter of the rectangle is 45 inches.

Steps:

Define the width as w. The length, L, is 4w. Use the perimeter formula: P 2L 2w. Substitute the values: 45 2(4w) 2w. Simplify: 45 8w 2w or 45 10w. Solve for w: w frac{45}{10} 4.5 inches. Calculate the length: l 4w 4 times 4.5 18 inches. Calculate the area: A l times w 18 times 4.5 or A 81 square inches.

Conclusion: The area of the rectangle is 81 square inches.

Example 3: Length is 3 Times the Width (Second Example)

Given:

The length of a rectangle is 3 times its width. The perimeter of the rectangle is 63 inches.

Steps:

Define the width as w. The length, L, is 3w. Use the perimeter formula: P 2L 2w. Substitute the values: 63 2(3w) 2w. Simplify: 63 6w 2w or 63 8w. Solve for w: w frac{63}{8} 7.875 inches. Calculate the length: l 3w 3 times 7.875 23.625 inches. Calculate the area: A l times w 23.625 times 7.875 or A approximately 186.328125 square inches.

Conclusion: The area of the rectangle is approximately 186.33 square inches.

Example 4: Length is 3 Times the Width (Third Example)

Given:

The length of a rectangle is 3 times its width.

Steps:

Define the width as w. The length, L, is 3w. Use the perimeter formula: P 2L 2w. Substitute the values: P 8w. Solve for w: 63 8w, thus w frac{63}{8} 7.875 inches. Calculate the length: l 3w 3 times 7.875 23.625 inches. Calculate the area: A l times w 23.625 times 7.875 or A approximately 186.328125 square inches.

Conclusion: The area of the rectangle is approximately 186.33 square inches.

Example 5: Length is 3 Times the Width (Fourth Example)

Given:

The length of a rectangle is 3 times its width. The perimeter of the rectangle is 63 inches.

Steps:

Define the width as w. The length, L, is 3w. Use the perimeter formula: P 2L 2w. Substitute the values: 63 8w. Solve for w: w frac{63}{8} 7.875 inches. Calculate the length: l 3w 3 times 7.875 23.625 inches. Calculate the area: A l times w 23.625 times 7.875 or A approximately 186.328125 square inches.

Conclusion: The area of the rectangle is approximately 186.33 square inches.