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A rectangular front porch has an area of 6060 square feet. Its perimeter is 3232 feet. What are the dimensions of the porch?\newline___\_\_\_ feet by ___\_\_\_ feet

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Q. A rectangular front porch has an area of 6060 square feet. Its perimeter is 3232 feet. What are the dimensions of the porch?\newline___\_\_\_ feet by ___\_\_\_ feet
  1. Define Area Equation: Let's denote the length of the porch as LL feet and the width as WW feet. We know that the area (AA) of a rectangle is given by the formula A=L×WA = L \times W. We are given that the area is 6060 square feet.\newlineSo, we have the equation:\newlineL×W=60L \times W = 60
  2. Define Perimeter Equation: We also know that the perimeter PP of a rectangle is given by the formula P=2L+2WP = 2L + 2W. We are given that the perimeter is 3232 feet.\newlineSo, we have the equation:\newline2L+2W=322L + 2W = 32
  3. Simplify Perimeter Equation: To find the dimensions LL and WW, we need to solve these two equations together. Let's simplify the perimeter equation by dividing all terms by 22 to make it easier to work with:\newlineL+W=16L + W = 16
  4. Express WW in Terms of LL: Now we have a system of two equations:\newline11) L×W=60L \times W = 60\newline22) L+W=16L + W = 16\newlineWe can solve this system by expressing one variable in terms of the other using the second equation. Let's express WW in terms of LL:\newlineW=16−LW = 16 - L
  5. Substitute WW into Area Equation: Substitute W=16−LW = 16 - L into the first equation (L×W=60L \times W = 60):\newlineL×(16−L)=60L \times (16 - L) = 60\newlineExpand the equation:\newline16L−L2=6016L - L^2 = 60
  6. Rearrange Equation to Quadratic: Rearrange the equation to form a quadratic equation: L2−16L+60=0L^2 - 16L + 60 = 0
  7. Solve Quadratic Equation: Now we need to solve the quadratic equation for LL. We can do this by factoring, completing the square, or using the quadratic formula. The equation looks like it can be factored:\newline(L−10)(L−6)=0(L - 10)(L - 6) = 0
  8. Find Possible Values for L: Setting each factor equal to zero gives us the possible values for L:\newlineL−10=0L - 10 = 0 or L−6=0L - 6 = 0\newlineSo, L=10L = 10 or L=6L = 6
  9. Calculate Possible Dimensions: If L=10L = 10, then W=16−L=16−10=6W = 16 - L = 16 - 10 = 6. If L=6L = 6, then W=16−L=16−6=10W = 16 - L = 16 - 6 = 10. Since a rectangle's length and width are interchangeable, we have two possible sets of dimensions that satisfy both the area and perimeter: 1010 feet by 66 feet or 66 feet by 1010 feet.

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