Problem 51 - Algebra and Trigonometry Real Mathematics, Real People

Question

A room is 1.5 times as long as it is wide, and its perimeter is 25 meters. (a) Draw a diagram that gives a visual representation of the problem. Identify the length as \(l\) and the width as \(w\) (b) Write \(l\) in terms of \(w\) and write an equation for the perimeter in terms of \(w\) (c) Find the dimensions of the room.

Step-by-Step Solution

Answer

The dimensions of the room are: length = 7.5m and width = 5m.

Step 1: Draw a Diagram

Start by drawing a rectangle to represent the rectangular room. Label one side as the width, \(w\), and the other as the length, \(l\). Note that \(l\) and \(w\) are the lengths of the side of the rectangle and not the area inside the rectangle.

Step 2: Write Length in Terms of Width

Since the length is 1.5 times the width, write \(l = 1.5w\).

Step 3: Write The Perimeter in Terms of Width

Next, substitute \(\)l from step 2 into the perimeter equation. So, the equation for the perimeter becomes \(P = 2(1.5w) + 2w = 25\).

Step 4: Solve for Width

Now, simplify the equation to \(w = 25 / 5 = 5\) meters.

Step 5: Find the Length

Substitute \(w = 5\) into \(l = 1.5w\) to get \(l = 1.5 * 5 = 7.5\) meters.

Key Concepts

Perimeter Calculation

Calculating the perimeter of a rectangle involves understanding that the perimeter is the total distance around the shape. For a rectangle, this is found by adding together the lengths of all four sides. You will need the formula:

• The perimeter, \(P\), is calculated as: \(P = 2l + 2w\)

where \(l\) represents the length and \(w\) represents the width of the rectangle. In our exercise, this equation becomes important because it helps us express the perimeter in terms of another variable, making solving simpler. In practice, once you've substituted numerical or variable values into this formula, it becomes a straightforward calculation to find any side if all others are known. As seen in the exercise, knowing the perimeter is 25 meters allows us to create an equation once the relation of width and length is known.

Rectangular Dimensions

Understanding rectangular dimensions means being able to relate the length and width based on given information. Often, these dimensions have a ratio or specific relationship. For example, in our room problem, it's provided that the room is 1.5 times as long as it is wide. This relationship is key:

• Let the width be \(w\).
• The length, \(l\), can be expressed in terms of the width as \(l = 1.5w\).

This approach transforms what we know into mathematical terms, allowing us to use substitution in equations. By turning the problem into manageable algebraic expressions, determination of one dimension helps in finding the corresponding dimension effortlessly. This is critical for solving dimension-related questions, ensuring clarity and simplifying the complication of multiple unknowns.

Equation Manipulation

Equation manipulation is a powerful algebraic tool that involves transforming given equations to isolate the unknowns or express them conveniently. In problems involving geometry and algebra, this can mean rewriting equations or substituting terms as needed.In the context of our problem:- We start with the perimeter equation: \(P = 2l + 2w\).- We substitute the length \(l\) from our rectangular dimension equation \(l = 1.5w\) into the perimeter equation.- This gives us: \(25 = 2(1.5w) + 2w\).The next step is simplification:- Combine like terms: \(2(1.5w) = 3w\) results in \(25 = 3w + 2w\), thus \(5w = 25\).- Solve for \(w\) by dividing both sides by 5, resulting in \(w = 5\).Equation manipulation helps transform the original problem into a form where the solution becomes transparent, revealing itself step-by-step as we adjust and simplify the expressions. By mastering different approaches to manipulating equations, one can find paths to solutions that weren't initially obvious.