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The perimeter of a rectangular shop in the mall is 3232 meters. The area is 6060 square meters. What are the dimensions of the shop?\newline___\_\_\_ meters by ___\_\_\_ meters

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Q. The perimeter of a rectangular shop in the mall is 3232 meters. The area is 6060 square meters. What are the dimensions of the shop?\newline___\_\_\_ meters by ___\_\_\_ meters
  1. Perimeter Equation: Let's denote the length of the shop as LL meters and the width as WW meters. The perimeter PP of a rectangle is given by the formula P=2(L+W)P = 2(L + W). We are given that the perimeter is 3232 meters.\newlineSo, we have the equation 2(L+W)=322(L + W) = 32.
  2. Simplify Perimeter Equation: We can simplify the equation by dividing both sides by 22 to find L+WL + W. \newlineL+W=322L + W = \frac{32}{2}\newlineL+W=16L + W = 16
  3. Area Equation: 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 meters.\newlineSo, we have the equation L×W=60L \times W = 60.
  4. Solve System of Equations: Now we have a system of two equations:\newline11) L+W=16L + W = 16\newline22) L×W=60L \times W = 60\newlineWe can solve this system of equations to find the values of LL and WW.
  5. Express WW in Terms of LL: Let's express WW in terms of LL from the first equation: W=16−LW = 16 - L. Now we can substitute WW in the second equation with (16−L)(16 - L). L×(16−L)=60L \times (16 - L) = 60
  6. Expand and Rearrange Equation: Expanding the equation, we get:\newline16L−L2=6016L - L^2 = 60\newlineRearranging the terms, we get a quadratic equation:\newlineL2−16L+60=0L^2 - 16L + 60 = 0
  7. Solve Quadratic Equation: We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. The equation factors nicely as: \newline(L−10)(L−6)=0(L - 10)(L - 6) = 0
  8. Factor Quadratic Equation: Setting each factor equal to zero gives us the possible values for LL:L−10=0L - 10 = 0 or L−6=0L - 6 = 0So, L=10L = 10 or L=6L = 6
  9. Find Possible Values for LL: 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. Both pairs (L=10L = 10, W=6W = 6) and (L=6L = 6, W=10W = 10) are valid solutions since they are interchangeable (length and width can be swapped).

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