Problem 43 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The length and width are 20 and 15, respectively.

Step 1: Understand the Problem

The perimeter of a rectangle is given by the formula: \( P = 2(l + w) \), where \( l \) is the length and \( w \) is the width. The diagonal of a rectangle can be found using the Pythagorean theorem: \( d = \sqrt{l^2 + w^2} \). We are given \( P = 70 \) and \( d = 25 \). We need to find \( l \) and \( w \).

Step 2: Set Up the Equations

From the perimeter formula, we have:\[2(l + w) = 70 \]From this, we can simplify to:\[l + w = 35 \]From the diagonal formula: \[l^2 + w^2 = 25^2 \]which simplifies to:\[l^2 + w^2 = 625 \]

Step 3: Express Width in Terms of Length

From the equation \( l + w = 35 \), we can express \( w \) as:\[w = 35 - l \]

Step 4: Substitute and Solve for Length

Substitute \( w = 35 - l \) into the diagonal equation:\[l^2 + (35 - l)^2 = 625 \]Expand and simplify:\[l^2 + 1225 - 70l + l^2 = 625 \]Combine like terms:\[2l^2 - 70l + 1225 = 625 \]Subtract 625 from both sides:\[2l^2 - 70l + 600 = 0 \]Divide the entire equation by 2:\[l^2 - 35l + 300 = 0 \]

Step 5: Solve the Quadratic Equation

The quadratic equation \( l^2 - 35l + 300 = 0 \) can be solved using the quadratic formula:\[l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \( a = 1 \), \( b = -35 \), and \( c = 300 \):\[l = \frac{-(-35) \pm \sqrt{(-35)^2 - 4 \times 1 \times 300}}{2 \times 1} \]\[l = \frac{35 \pm \sqrt{1225 - 1200}}{2} \]\[l = \frac{35 \pm \sqrt{25}}{2} \]\[l = \frac{35 \pm 5}{2} \]This yields two solutions for \( l \):1. \( l = \frac{40}{2} = 20 \)2. \( l = \frac{30}{2} = 15 \)

Step 6: Determine the Dimensions

Using the solutions from the quadratic equation:1. If \( l = 20 \), then \( w = 35 - 20 = 15 \)2. If \( l = 15 \), then \( w = 35 - 15 = 20 \)Thus, the length is 20 and width is 15, or vice versa, since rectangle dimensions are interchangeable.

Key Concepts

Perimeter

When you think of a rectangle, imagine it as a shape with four sides. The perimeter is the distance all around this shape. A rectangle has two lengths and two widths. These sides are quite important in calculating the perimeter. The formula is simple: add up the lengths and the widths and then multiply by two.
The formula goes:

• The perimeter, \( P = 2(l + w) \)
• Here, \( l \) is the length and \( w \) is the width.

Given a perimeter, we can find the sum of the length and the width. For example, if you know the perimeter is 70, you'll set up this equation: \( 2(l + w) = 70 \). Simplify it to \( l + w = 35 \). This equation will help us in figuring out how long and wide the rectangle is, which is a key step in solving the problem.

Pythagorean theorem

The Pythagorean theorem is magical for right-angled triangles, like the ones you find in rectangles when you split their diagonals. Recall this powerful rule: in a right triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides.
For a rectangle, the diagonal forms such a triangle with the length and the width as the two other sides. You can write the Pythagorean theorem as:

• \( d^2 = l^2 + w^2 \)
• Where \( d \) is the diagonal, \( l \) is the length, and \( w \) is the width.

In our rectangle with a diagonal of 25, this becomes \( 25^2 = l^2 + w^2 \) or \( 625 = l^2 + w^2 \). This equation helps us express a connection between the length and the width. With it, we can solve for one measurement if we know the other, helping us find the rectangle's complete dimensions.

Quadratic equation

Quadratic equations might sound tricky, but they're quite handy when dealing with areas involving squared terms. A typical quadratic follows the structure: \( ax^2 + bx + c = 0 \).
In our rectangle problem, we transformed the equations into a quadratic by using expressions for length and width. Following simplification, we ended up with:

• \( l^2 - 35l + 300 = 0 \)

This is where the quadratic formula jumps in to save the day. To solve it, use:

• \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Substitute values: \( a = 1 \), \( b = -35 \), \( c = 300 \)
• Calculate: \( l = \frac{35 \pm \sqrt{1225 - 1200}}{2} \)

Once calculated, this gives us two possible answers for \( l \): 15 and 20. These solutions match up with the dimensions of the rectangle, accounting for both length and width interchangeably.