Math Problem Statement

The area of a rectangle is52yd^2, and the length of the rectangle is 5yd less than twice the width. Find the dimensions of the rectangle.

Solution

Let the width of the rectangle be xx yards.
Then, the length of the rectangle is given as 5 yards less than twice the width, which can be expressed as:
Length=2x5\text{Length} = 2x - 5

The area of the rectangle is given as 52 square yards, so we use the formula for the area of a rectangle:

Area=Length×Width\text{Area} = \text{Length} \times \text{Width}

Substituting the given values:

x(2x5)=52x(2x - 5) = 52

Step 1: Expand the equation

2x25x=522x^2 - 5x = 52

Step 2: Rearrange into standard quadratic form

2x25x52=02x^2 - 5x - 52 = 0

Step 3: Solve for xx using the quadratic formula

The quadratic formula is:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For the equation 2x25x52=02x^2 - 5x - 52 = 0, we have:

  • a=2a = 2
  • b=5b = -5
  • c=52c = -52

First, compute the discriminant:

b24ac=(5)24(2)(52)b^2 - 4ac = (-5)^2 - 4(2)(-52) =25+416=441= 25 + 416 = 441

Now, take the square root:

441=21\sqrt{441} = 21

Now, solve for xx:

x=(5)±212(2)x = \frac{-(-5) \pm 21}{2(2)}

x=5±214x = \frac{5 \pm 21}{4}

Step 4: Find possible values for xx

x=5+214=264=6.5x = \frac{5 + 21}{4} = \frac{26}{4} = 6.5

x=5214=164=4x = \frac{5 - 21}{4} = \frac{-16}{4} = -4

Since the width cannot be negative, we take:

x=6.5 yards (width)x = 6.5 \text{ yards (width)}

Step 5: Find the length

Length=2(6.5)5=135=8 yards\text{Length} = 2(6.5) - 5 = 13 - 5 = 8 \text{ yards}

Final Answer:

  • Width = 6.5 yards
  • Length = 8 yards

Would you like a more detailed breakdown or verification of the steps?


Related Questions:

  1. How do you solve quadratic equations by factoring instead of using the quadratic formula?
  2. What are the properties of a rectangle that distinguish it from other quadrilaterals?
  3. How can you determine the dimensions of a rectangle if given perimeter instead of area?
  4. How does changing the width affect the area of a rectangle while keeping the length proportional?
  5. What real-world applications involve finding dimensions of a rectangle given area constraints?

Math Tip:

Always check whether the quadratic equation can be factored easily before using the quadratic formula, as factoring can save time. 🚀

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Geometry

Formulas

Area of a rectangle = Length × Width
Quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Quadratic formula

Suitable Grade Level

Grades 7-9