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The area of a rectangular piece of metal is 36 square centimeters36\,\text{square centimeters}. The perimeter is 26 centimeters26\,\text{centimeters}. What are the dimensions of the piece?\newline____\_\_\_\_ centimeters by ____\_\_\_\_ centimeters

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Q. The area of a rectangular piece of metal is 36 square centimeters36\,\text{square centimeters}. The perimeter is 26 centimeters26\,\text{centimeters}. What are the dimensions of the piece?\newline____\_\_\_\_ centimeters by ____\_\_\_\_ centimeters
  1. Define Variables: Let's denote the length of the rectangle as LL cm and the width as WW cm. We are given two equations based on the area and perimeter of the rectangle:\newline11. Area (A)=L×W(A) = L \times W\newline22. Perimeter (P)=2×(L+W)(P) = 2 \times (L + W)\newlineFrom the problem, we know:\newlineA=36 cm2A = 36 \, \text{cm}^2\newlineP=26 cmP = 26 \, \text{cm}\newlineWe can use these two equations to find the values of LL and WW.
  2. Write Equations: First, let's write down the equations with the given values:\newline11. L×W=36L \times W = 36\newline22. 2×(L+W)=262 \times (L + W) = 26\newlineNow, we can simplify the second equation to find an expression for L+WL + W:\newline2×(L+W)=262 \times (L + W) = 26\newlineL+W=262L + W = \frac{26}{2}\newlineL+W=13L + W = 13
  3. Simplify Equations: We have two equations now:\newline11. L×W=36L \times W = 36\newline22. L+W=13L + W = 13\newlineWe can use substitution or elimination to solve these equations. Let's express WW in terms of LL using the second equation:\newlineW=13−LW = 13 - L
  4. Substitute and Solve: Now, we substitute W=13−LW = 13 - L into the first equation:\newlineL×(13−L)=36L \times (13 - L) = 36\newlineExpanding this, we get a quadratic equation:\newlineL2−13L+36=0L^2 - 13L + 36 = 0\newlineTo find the values of LL, we need to factor this quadratic equation.
  5. Factor Quadratic Equation: Factoring the quadratic equation L2−13L+36=0L^2 - 13L + 36 = 0, we look for two numbers that multiply to 3636 and add up to 1313. These numbers are 99 and 44.\newlineSo, the factors are:\newline(L−9)(L−4)=0(L - 9)(L - 4) = 0\newlineSetting each factor equal to zero gives us the possible values for LL:\newlineL−9=0L - 9 = 0 or L−4=0L - 4 = 0\newlineL=9L = 9 or 363600
  6. Find Possible Values: Since LL can be either 99 or 44, we need to determine the corresponding width for each case. If L=9L = 9 cm, then W=13−L=13−9=4W = 13 - L = 13 - 9 = 4 cm. If L=4L = 4 cm, then W=13−L=13−4=9W = 13 - L = 13 - 4 = 9 cm.\newlineTherefore, the dimensions of the rectangle can be either 99 cm by 44 cm or 44 cm by 99 cm. Both sets of dimensions satisfy the area and perimeter conditions given in the problem.

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