Solving for the Dimensions of a Rectangle with a Given Perimeter

In this article, we will guide you through solving a problem involving the dimensions of a rectangle when the perimeter is given. Specifically, we will solve the problem where the width of the rectangle is half of its length, and the perimeter is 63 inches. By following a step-by-step approach, we will determine the length and width of the rectangle.

Problem Statement

The width of a rectangle is 1/2 of its length. If its perimeter is 63 inches, what are the sides of the rectangle?

Step-by-Step Solution

Let the length of the rectangle be L inches and the width be W inches. According to the problem, the width is half of the length:

W 1/2 L

The formula for the perimeter P of a rectangle is given by:

P 2L 2W

Substituting the expression for W into the perimeter formula, we get:

P 2L 2(1/2 L) 2L L 3L

We know that the perimeter is 63 inches, so we can set up the equation:

3L 63

Solving for L:

L 63/3 21 inches

Substituting back to find W:

W 1/2 L 1/2 * 21 10.5 inches

Therefore, the dimensions of the rectangle are:

Length L 21 inches

Width W 10.5 inches

Alternative Solution

Another approach involves considering the relationship between the length and width directly:

L - L/2 63/2 or 3L 63

L 63/3 21 inches and W 1/2 L 1/2 * 21 10.5 inches

Hence, the length is 21 inches and the width is 10.5 inches.

Generalized Formulation

Let us assume the length to be 2x, then the width will be x (since the width is 1/2 of the length).

Perimeter of Rectangle 2 (length width) 2 (2x x) 63

63 2 (3x/2)

x 63/6 10.5 inches

Hence, the width is 10.5 inches, and the length is 2x 21 inches.

Another Approach

Let the length be x, so the width is 1/2 of its length:

Width x/2

We know that the perimeter of a rectangle is (2 times (length times width)), and according to the question:

2x x/2 63

2 (2x x/2) 63

2 (3x/2) 63

3x 63

x 63/3 21

Therefore, the length is 21 inches, and the width is 21/2 10.5 inches.