Ask Question
11 November, 22:35

the area of a rectangle is 66 yd^2, and the length of the rectangle is 1 yd more than twice the width. find the dimensions of the rectangle.

+1
Answers (1)
  1. 11 November, 22:38
    0
    The length of the rectangle is 11 yards, the width of the rectangle is 6 yards.

    Step-by-step explanation:

    1. Let's check the information given to resolve the question:

    Width of the rectangle = x

    Length of the rectangle = 2x - 1 (Length of the rectangle is 1 yd more than twice the width)

    Area of the rectangle = 66 square yards

    2. Let's find the value of the length and the width

    Area of the rectangle = Length of the rectangle * Width of the rectangle

    66 = x (2x - 1)

    66 = 2x² - x

    0 = 2x² - x - 66 (Subtracting 66 at both sides)

    This is a quadratic equation and the formula to solve it is this:

    x = [ - b + / - √ (b² - 4ac) / 2a]

    x = - [ - ( - 1) + / - √ (1 ² - 4 (2) (-66)) / 2 (2) ]

    x = - [1 + / - √ (1 + 528) / 4]

    x = - [1 + / - √529) / 4]

    x = - [1 + / - 23/4]

    x₁ = 24/4 = 6

    x₂ = - 22/4 = - 5.5

    We will take the value of x₁ because x₂ is negative and the width can't be negative.

    Width of the rectangle = 6 yards

    Length of the rectangle = 2x - 1 = 2 (6) - 1 = 12 - 1 = 11 yards

    The length of the rectangle is 11 yards, the width of the rectangle is 6 yards.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “the area of a rectangle is 66 yd^2, and the length of the rectangle is 1 yd more than twice the width. find the dimensions of the rectangle. ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers