# A rectangular patio is 8 feet longer than it is wide. The area of the patio is 84 square feet....

## Question:

A rectangular patio is 8 feet longer than it is wide. The area of the patio is 84 square feet. What are the dimensions of the patio?

## Rectangles:

A rectangle is a quadrilateral with four sides. Each of the four interior angles of a rectangle is 90 degrees. An example of a figure with a rectangular shape is a 10 by 12 picture frame.

The area of a rectangle is given by:

• {eq}A = lw {/eq}

Let the width of the rectangular patio be:

• {eq}w = x\, \rm ft {/eq}

If the length is 8ft longer than the width, the length can be expressed as:

• {eq}l = (x + 8)\, \rm ft {/eq}

The area of the patio is equal to:

• {eq}A = x(x + 8) {/eq}
• {eq}A = x^2 + 8x {/eq}

If the area of patio is 84 square feet, we can write:

• {eq}84 = x^2 + 8x {/eq}
• {eq}x^2 + 8x - 84 = 0 {/eq}

We have a quadratic equation to solve. We'll solve the equation using the quadratic formula.

• {eq}x = \dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a} {/eq}

From our quadratic equation, we have:

• {eq}a = 1 {/eq}
• {eq}b = 8 {/eq}
• {eq}c = 84 {/eq}
• {eq}x = \dfrac{-8\pm \sqrt{8^2 - 4\times 1\times -84}}{2} {/eq}
• {eq}x = \dfrac{-8\pm \sqrt{64 + 340}}{2} {/eq}
• {eq}x = \dfrac{-8\pm 20}{2} {/eq}
• {eq}x = 6, \quad x = -14 {/eq}

Since length cannot be negative, we will consider the positive value of x.

Thus, the width of the patio is:

• {eq}w = x = \boxed{6\, \rm ft} {/eq}

And the length is equal to:

• {eq}l = 6 +8 = \boxed{14\, \rm ft} {/eq}