Problem 98 - Elementary Algebra

Question

The length of a rectangle is 3 times its width. If the diagonal measures 2 feet, then find the dimensions of the rectangle.

Step-by-Step Solution

Answer

Width: \( \sqrt{\frac{2}{5}} \), Length: \( 3 \times \sqrt{\frac{2}{5}} \).

Step 1: Understand the Problem

We need to find the dimensions of a rectangle, where the length is 3 times the width and the diagonal is 2 feet. We'll use these conditions to set up an equation.

Step 2: Set Up Variables

Let's denote the width of the rectangle as \( w \). According to the problem, the length \( l \) is 3 times the width, so \( l = 3w \).

Step 3: Use the Pythagorean Theorem

Since the rectangle's diagonal, width, and length form a right triangle, we can use the Pythagorean theorem: \[ l^2 + w^2 = d^2 \] where \( d = 2 \) is the diagonal.

Step 4: Substitute and Simplify

Substitute \( l = 3w \) and \( d = 2 \) into the equation. This gives us: \[ (3w)^2 + w^2 = 2^2 \]. Simplify to get \[ 9w^2 + w^2 = 4 \].

Step 5: Combine Like Terms

Combine the terms on the left side to get: \[ 10w^2 = 4 \].

Step 6: Solve for \( w \)

Divide both sides by 10 to isolate \( w^2 \): \[ w^2 = \frac{4}{10} \], which simplifies to \[ w^2 = \frac{2}{5} \]. Then take the square root of both sides: \[ w = \sqrt{\frac{2}{5}} \].

Step 7: Calculate \( l \)

Now, use \( l = 3w \) to find the length: \( l = 3 \times \sqrt{\frac{2}{5}} \).

Step 8: Simplify and Verify

Simplify the expressions \( w = \sqrt{\frac{2}{5}} \) and \( l = 3 \times \sqrt{\frac{2}{5}} \) to ensure they satisfy the original diagonal condition when squared and summed: \[ (") (\sqrt{\frac{2}{5}})^2 + (3 \times \sqrt{\frac{2}{5}})^2 = 4 \]. Confirm the equation holds true.

Key Concepts

Rectangle Dimensions

Understanding the dimensions of a rectangle is essential in geometry. A rectangle is defined by its four sides, with opposite sides being equal in length and adjacent sides forming right angles. In this problem, the rectangle's length is described as being three times its width. To determine these dimensions, we assign a variable to represent the width, often denoted as \( w \). If the length \( l \) is three times the width, our equation becomes \( l = 3w \).

Finding these sides is crucial because they frame the rectangle, allowing us to calculate other properties, such as perimeter and area. For instance:

• The perimeter \( P \) of a rectangle is given by \( P = 2l + 2w \).
• The area \( A \) of a rectangle is found by \( A = lw \).

By understanding these foundational properties, you can unlock various geometric applications and solve complex problems, such as those involving ratios as seen here.

Right Triangle

A right triangle is a triangle in which one angle measures 90 degrees. This specific type of triangle is significant in the context of rectangles. When you draw a diagonal across a rectangle, it divides the rectangle into two right triangles. This introduces a perfect opportunity to utilize the Pythagorean Theorem.

The Pythagorean Theorem is stated as \( a^2 + b^2 = c^2 \), where \( a \) and \( b \) are the legs of the triangle, and \( c \) is the hypotenuse (the longest side of the triangle). In the case of our rectangle, the length \( l \) and width \( w \) serve as the legs of our right triangle, and the diagonal \( d \) is the hypotenuse. Thus, the equation becomes \( l^2 + w^2 = d^2 \).

This relation helps in proving many geometric principles and solving problems that involve the properties and relations of shapes.

Diagonal Measurement

The diagonal of a rectangle acts as a critical line segment within the shape, providing insight into its dimensions through geometric principles. In our given problem, the diagonal measures 2 feet, serving as a crucial data point.

A diagonal splits a rectangle into two equal right triangles and can be calculated using the Pythagorean Theorem: \( l^2 + w^2 = d^2 \). Substitute the known value of the diagonal (2 feet) into this equation to solve it stepwise, finding that the dimensions meet this condition:

• For the rectangle with an unknown width \( w \) and a length three times that width, we substitute \( l = 3w \) into the theorem.
• This results in \( 9w^2 + w^2 = 4 \), simplifying to \( 10w^2 = 4 \).
• Solve for \( w \) and subsequently use \( l = 3w \) to find the length.

This process highlights how a single diagonal measurement can help determine the entire framework of the rectangle, illustrating the power of geometry in solving real-world problems.