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The perimeter of an envelope is 2828 inches. The area is 4040 square inches. What are the dimensions of the envelope?\newline____\_\_\_\_ inches by ____\_\_\_\_ inches

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Q. The perimeter of an envelope is 2828 inches. The area is 4040 square inches. What are the dimensions of the envelope?\newline____\_\_\_\_ inches by ____\_\_\_\_ inches
  1. Define Variables: Let's denote the length of the envelope as ll inches and the width as ww inches.\newlineThe perimeter (PP) of a rectangle is given by the formula P=2l+2wP = 2l + 2w.
  2. Perimeter Equation: Given the perimeter of the envelope is 2828 inches, we can write the equation 2l+2w=282l + 2w = 28.
  3. Simplify Equation: We can simplify this equation by dividing all terms by 22, which gives us l+w=14l + w = 14.
  4. Area Formula: The area AA of a rectangle is given by the formula A=lwA = lw.
  5. Area Equation: Given the area of the envelope is 4040 square inches, we can write the equation lw=40lw = 40.
  6. Solve System of Equations: Now we have a system of two equations with two variables:\newline11. l+w=14l + w = 14\newline22. lw=40lw = 40\newlineWe can solve this system by expressing one variable in terms of the other using the first equation and then substituting into the second equation.
  7. Express ww in terms of ll: Let's express ww in terms of ll from the first equation: w=14lw = 14 - l.
  8. Substitute into Equation: Substitute w=14lw = 14 - l into the second equation lw=40lw = 40:l(14l)=40l(14 - l) = 40
  9. Expand Equation: Expand the equation: 14ll2=4014l - l^2 = 40.
  10. Rearrange Equation: Rearrange the equation to form a quadratic equation: l214l+40=0l^2 - 14l + 40 = 0.
  11. Factor Quadratic Equation: Factor the quadratic equation: l - \(10)(l - 44) = 00\
  12. Solve for ll: Set each factor equal to zero and solve for ll:1.1. l10=0l - 10 = 0 gives l=10l = 102.2. l4=0l - 4 = 0 gives l=4l = 4
  13. Find Solutions: Since ll and ww are interchangeable as length and width, we can have two solutions:\newline11. l=10l = 10 inches and w=4w = 4 inches\newline22. l=4l = 4 inches and w=10w = 10 inches\newlineBoth sets of dimensions satisfy the given perimeter and area.

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