Solving for the Dimensions of a Rectangle: A Mathematical Puzzle

Introduction

Mathematics often presents us with intriguing puzzles, particularly in the field of geometry. One such puzzle involves finding the dimensions of a rectangle based on certain conditions related to its width, length, and perimeter. Let's explore this problem step by step.

The Problem Statement

We are given the following conditions for a rectangle:

The length of a rectangle is 12 units more than its width. The perimeter of the rectangle is 7 times the width.

Formulating the Equations

We can represent the width of the rectangle as w and the length as l. Thus, we can write the following equations based on the given conditions:

Length (l) in terms of width (w): l w 12 Perimeter (P) in terms of width (w): P 2l 2w Perimeter (P) also equals 7 times the width: P 7w

Setting up and Solving the Equations

Let's combine the equations to find the width and length of the rectangle.

Substitute the expression for l from the first equation into the perimeter equation: 2(w 12) 2w 7w Expand and simplify: 2w 24 2w 7w 4w 24 7w Isolate w: 24 7w - 4w 24 3w Solve for w: w 8

Now that we have the width, we can find the length:

l w 12 8 12 20 Thus, the length of the rectangle is 20 units.

Verifying the Solution

Let's verify the solution using the derived values:

Width: w 8 units

Length: l 20 units

Perimeter: P 2l 2w 2(20) 2(8) 40 16 56 units

The perimeter should also be equal to 7 times the width:

7w 7(8) 56 units

Since both expressions equal 56 units, our solution is correct.

Summary

Through algebraic manipulation and solving equations, we have determined that the width of the rectangle is 8 units and the length is 20 units. This problem showcases the power of algebra in solving geometric puzzles.