The problem states that a rectangle's length is 10 feet longer than its width. The area of the rectangle is 75 square feet. We need to find the dimensions (length and width) of the rectangle.

1. Problem Description

The problem states that a rectangle's length is 10 feet longer than its width. The area of the rectangle is 75 square feet. We need to find the dimensions (length and width) of the rectangle.

2. Solution Steps

Let $w$ be the width of the rectangle in feet.

Since the length is 10 feet longer than the width, the length $l$ is given by:

$l = w + 10$

The area of a rectangle is given by the formula:

$Area = length \times width$

$A = l \times w$

We are given that the area is 75 square feet. So, we have:

$75 = (w + 10) \times w$

$75 = w^2 + 10w$

Rearrange the equation to form a quadratic equation:

$w^2 + 10w - 75 = 0$

Now we solve the quadratic equation for $w$. We can factor the quadratic equation:

$(w + 15)(w - 5) = 0$

This gives us two possible solutions for $w$:

$w + 15 = 0 \Rightarrow w = -15$

$w - 5 = 0 \Rightarrow w = 5$

Since the width cannot be negative, we discard the solution $w = -15$. Therefore, the width is $w = 5$ feet.

Now we find the length:

$l = w + 10 = 5 + 10 = 15$ feet.

So the dimensions of the rectangle are 5 feet by 15 feet. We can check our answer by calculating the area:

$Area = l \times w = 15 \times 5 = 75$ square feet. This confirms our solution.

3. Final Answer

The dimensions of the rectangle are width = 5 feet and length = 15 feet.