Problem 47 - Precalculus: Mathematics for Calculus

Question

The perimeter of a rectangle is \(70,\) and its diagonal is \(25 .\) Find its length and width.

Step-by-Step Solution

Answer

Length is 20 and width is 15, or length is 15 and width is 20.

Step 1: Understand the Problem

We need to find the length and width of a rectangle given that its perimeter is 70 and its diagonal is 25. We have two unknowns, length (\(l\)) and width (\(w\)).

Step 2: Set Up the Perimeter Equation

The formula for the perimeter of a rectangle is \(P = 2l + 2w\). Given that the perimeter \(P = 70\), we can write the equation: \[ 2l + 2w = 70 \] which simplifies to: \[ l + w = 35. \]

Step 3: Set Up the Diagonal Equation

The formula for the diagonal \(d\) of a rectangle is \(d = \sqrt{l^2 + w^2}\). Given that the diagonal \(d = 25\), we have: \[ \sqrt{l^2 + w^2} = 25. \] Squaring both sides gives: \[ l^2 + w^2 = 625. \]

Step 4: Solve the System of Equations

We now have two equations: (1) \(l + w = 35\) and (2) \(l^2 + w^2 = 625\).First, solve equation (1) for \(l\):\[ l = 35 - w. \]Substitute \(l = 35 - w\) into equation (2):\[ (35 - w)^2 + w^2 = 625. \]Expanding \((35 - w)^2\), we get:\[ 1225 - 70w + w^2 + w^2 = 625. \]Combine like terms:\[ 2w^2 - 70w + 1225 = 625. \]

Step 5: Simplify and Solve Quadratic Equation

First, subtract 625 from both sides:\[ 2w^2 - 70w + 600 = 0. \]Divide the entire equation by 2:\[ w^2 - 35w + 300 = 0. \]To solve this equation, we can either factor it or use the quadratic formula. Factoring it gives:\[ (w - 15)(w - 20) = 0. \]So, \(w = 15\) or \(w = 20\).

Step 6: Determine Length and Width

Plug the values of \(w\) back into \(l = 35 - w\):- If \(w = 15\), then \(l = 35 - 15 = 20\).- If \(w = 20\), then \(l = 35 - 20 = 15\).These give possible dimensions of the rectangle: length = 20, width = 15, or length = 15, width = 20.

Key Concepts

Perimeter of a Rectangle

A rectangle is a four-sided shape with opposite sides that are equal in length. Understanding the perimeter is an important concept in solving rectangle problems. The perimeter is the total distance around the rectangle. To calculate it, you add up all the sides. Since a rectangle has two pairs of equal sides, the formula is:

\[ P = 2l + 2w \]
Where:

• \( P \) is the perimeter.
• \( l \) is the length.
• \( w \) is the width.

For example, if a rectangle has a length of 10 and a width of 5, its perimeter is \( 2(10) + 2(5) = 20 + 10 = 30 \). In our specific problem, the perimeter is 70, which leads to the equation:\[ l + w = 35 \]
These relationships are vital for finding unknown dimensions of a rectangle.

Diagonal of a Rectangle

The diagonal of a rectangle is the straight line connecting opposite corners. Calculating the diagonal requires understanding the Pythagorean theorem because a rectangle’s diagonal forms a right triangle with its sides.

The theorem states that in a right triangle, the square of the hypotenuse (diagonal, in this case) equals the sum of the squares of the other two sides. The formula is:

\[ d = \sqrt{l^2 + w^2} \]
Where:

• \( d \) is the diagonal.
• \( l \) is the length.
• \( w \) is the width.

This formula helps find a diagonal when the side lengths are known. If the diagonal and one side are known, you can solve for the other side. In our exercise, knowing the diagonal as 25 gave us another equation:\[ l^2 + w^2 = 625 \]

Solving Systems of Equations

To find both unknown length and width of the rectangle, we need to solve a system of equations. In problems like this, you often end up with two equations that must be solved together, helping you find the values of two unknowns.

Systems of equations can be solved using the substitution method, elimination method, or graphing. In this exercise, we used substitution by expressing one variable in terms of another using one equation:
\[ l = 35 - w \]
Then, substituting into the second equation to solve for one variable:
\[ (35 - w)^2 + w^2 = 625 \]
This makes the problem more straightforward by reducing the two variables to one, allowing us to solve using simpler algebraic techniques.

Quadratic Equations

Quadratic equations are commonly used in geometrical problems like finding the dimensions of rectangles when you have squared terms. A standard form of a quadratic equation is:

\[ ax^2 + bx + c = 0 \]
For this problem, after substitution and simplification, we ended up with:
\[ w^2 - 35w + 300 = 0 \]
Solving quadratic equations can be done by:

• Factoring, as shown in the solution by expressing as product of binomials: \[ (w - 15)(w - 20) = 0 \]
• Using the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Quadratics reveal the possible values for the variable which, in this case, are the potential lengths and widths of the rectangle. For the system, the result was that the width could be 15 or 20, leading to corresponding lengths due to the substituted equation.