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The length of an 8000 square foot rectangular gymnasium is 20 feet greater than its width. What is its width, in feet?

The length of an 80008000 square foot rectangular gymnasium is 2020 feet greater than its width. What is its width, in feet?

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Full solution

Q. The length of an 80008000 square foot rectangular gymnasium is 2020 feet greater than its width. What is its width, in feet?
  1. Define Variables: Let's denote the width of the gymnasium as ww (in feet). According to the problem, the length is 2020 feet greater than the width, so we can denote the length as w+20w + 20 (in feet). The area of a rectangle is given by the formula length ×\times width. We are given that the area of the gymnasium is 80008000 square feet. So, we can set up the equation:\newlineArea = width ×\times length\newline8000=w×(w+20)8000 = w \times (w + 20)
  2. Expand Equation: Now we need to solve the quadratic equation for ww. First, we expand the right side of the equation:\newline8000=w2+20w8000 = w^2 + 20w
  3. Set to Zero: To solve the quadratic equation, we need to set it to zero by moving all terms to one side:\newlinew2+20w−8000=0w^2 + 20w - 8000 = 0
  4. Apply Quadratic Formula: We can attempt to factor the quadratic equation, but it does not factor nicely. Therefore, we will use the quadratic formula to solve for ww. The quadratic formula is given by:\newlinew=−b±b2−4ac2aw = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\newlineIn our equation, a=1a = 1, b=20b = 20, and c=−8000c = -8000.
  5. Substitute Values: We substitute the values of aa, bb, and cc into the quadratic formula:\newlinew=−20±202−4(1)(−8000)2(1)w = \frac{-20 \pm \sqrt{20^2 - 4(1)(-8000)}}{2(1)}\newlinew=−20±400+320002w = \frac{-20 \pm \sqrt{400 + 32000}}{2}\newlinew=−20±324002w = \frac{-20 \pm \sqrt{32400}}{2}
  6. Calculate Discriminant: We calculate the discriminant (32400\sqrt{32400}) and simplify the expression:\newlinew=−20±1802w = \frac{-20 \pm 180}{2}
  7. Find Positive Solution: Since we are looking for a positive width, we take the positive solution of the quadratic formula:\newlinew=(−20+180)/2w = (-20 + 180) / 2\newlinew=160/2w = 160 / 2\newlinew=80w = 80
  8. Check Solution: We have found the width of the gymnasium to be 8080 feet. We should check that this width gives us the correct area when multiplied by the length (which is 2020 feet greater than the width).\newlineLength = w+20=80+20=100w + 20 = 80 + 20 = 100 feet\newlineArea = width ×\times length = 8080 feet ×\times 100100 feet = 80008000 square feet\newlineThis matches the given area, so our solution is correct.

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