Math Problem Statement

Pools Ever is building a new community pool in Harlem. The pool is surrounded by a 2-foot wide walkway. The width of the pool is 23 feet less than the length of the pool. The total area enclosed by the outer edge of the walkway is 1674 ft?. What are the dimensions of the pool?

Solution

To find the dimensions of the pool, let's break down the problem.

Step 1: Define the Variables

  • Let the length of the pool be LL feet.
  • Let the width of the pool be WW feet.

According to the problem, the width is 23 feet less than the length: W=L23W = L - 23

Step 2: Consider the Walkway

The walkway is 2 feet wide, so the total dimensions (including the pool and walkway) would be:

  • The total length: L+4L + 4 feet (2 feet on each side)
  • The total width: W+4W + 4 feet (2 feet on each side)

Step 3: Write the Equation for the Total Area

The total area enclosed by the outer edge of the walkway is given as 1674 square feet: (L+4)×(W+4)=1674(L + 4) \times (W + 4) = 1674

Step 4: Substitute the Expression for WW

Substitute W=L23W = L - 23 into the equation: (L+4)×((L23)+4)=1674(L + 4) \times ((L - 23) + 4) = 1674 Simplify the equation: (L+4)×(L19)=1674(L + 4) \times (L - 19) = 1674

Step 5: Expand and Solve the Quadratic Equation

Expand the equation: L219L+4L76=1674L^2 - 19L + 4L - 76 = 1674 L215L1750=0L^2 - 15L - 1750 = 0

Step 6: Solve the Quadratic Equation

Now, solve the quadratic equation L215L1750=0L^2 - 15L - 1750 = 0 using the quadratic formula: L=b±b24ac2aL = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=15b = -15, and c=1750c = -1750.

L=(15)±(15)24(1)(1750)2(1)L = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(-1750)}}{2(1)} L=15±225+70002L = \frac{15 \pm \sqrt{225 + 7000}}{2} L=15±72252L = \frac{15 \pm \sqrt{7225}}{2} L=15±852L = \frac{15 \pm 85}{2}

This gives us two possible solutions: L=1002=50orL=702=35L = \frac{100}{2} = 50 \quad \text{or} \quad L = \frac{-70}{2} = -35

Since the length cannot be negative, we have L=50L = 50 feet.

Step 7: Find the Width

Using W=L23W = L - 23: W=5023=27 feetW = 50 - 23 = 27 \text{ feet}

Conclusion:

The dimensions of the pool are:

  • Length: 50 feet
  • Width: 27 feet

Would you like more details or have any other questions?


Here are some related questions you might find interesting:

  1. How would the dimensions change if the walkway were 3 feet wide instead of 2 feet?
  2. What would be the total area of the pool itself without the walkway?
  3. How do you derive the quadratic formula from the general quadratic equation?
  4. Can you calculate the perimeter of the pool including the walkway?
  5. What is the difference in area between the pool and the walkway?
  6. How does the discriminant of a quadratic equation determine the nature of the roots?
  7. If the width was decreased by 5 feet, what would be the new length and area?
  8. Can this problem be solved using a different method, such as completing the square?

Tip: When dealing with quadratic equations, always check if the discriminant (the part under the square root) is non-negative to ensure real solutions.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Area Calculations
Linear Equations

Formulas

Quadratic formula

Theorems

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Suitable Grade Level

Grades 7-9