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The perimeter of a rectangle is a measurement that shows the total length around the rectangle. It's calculated by adding all the sides together. Since a rectangle has two pairs of equal sides, we can take the length, denoted by \( l \), and the width, denoted by \( w \), to calculate the perimeter with the formula

• \( P = 2(l + w) \)

In our problem, the perimeter is given as 70, so the equation becomes

• \( 2(l + w) = 70 \)

Simplifying this, we divide both sides by 2, resulting in

• \( l + w = 35 \)

This equation tells us that together, the length and the width must sum to 35. This forms one part of the system of equations used to solve for \( l \) and \( w \). Understanding how to find the perimeter is key to solving many geometric problems involving rectangles.

The Pythagorean theorem is an essential principle in geometry that relates the sides of a right triangle. In a rectangle, the diagonal creates two right triangles.

• The theorem states that the square of the length of the hypotenuse (longest side) is equal to the sum of the squares of the lengths of the other two sides.

Mathematically, it's expressed as

• \( a^2 + b^2 = c^2 \)

where \( a \) and \( b \) are the legs, and \( c \) is the hypotenuse.
For our rectangle problem, the diagonal \( d \) is the hypotenuse; therefore, we write

• \( l^2 + w^2 = d^2 \)

Given the diagonal \( d \) is 25, the equation becomes

• \( l^2 + w^2 = 625 \)

This equation is critical for solving the problem because it gives us another relationship between \( l \) and \( w \) allowing us to use both equations together to find the solution.

The quadratic formula is a powerful tool used to find the solutions of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). These equations often appear when solving geometric problems after expressing variables and simplifying them.

• The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our problem, after using the perimeter and Pythagorean equations, we derive the quadratic equation

• \( l^2 - 35l + 300 = 0 \)

Applying the quadratic formula involves identifying \( a = 1 \), \( b = -35 \), and \( c = 300 \).
First, we calculate the discriminant \( b^2 - 4ac \):

• \( (-35)^2 - 4 \times 1 \times 300 = 25 \).

With a positive discriminant, we find two solutions:

• \( l = \frac{35 + 5}{2} = 20 \)
• and \( l = \frac{35 - 5}{2} = 15 \).

This method effectively solves for \( l \), which in turn allows us to find \( w \) easily, completing the solution for both dimensions of the rectangle.