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The area of a rectangular ink pad is 48square centimeters48\,\text{square centimeters}. The perimeter is 28centimeters28\,\text{centimeters}. What are the dimensions of the ink pad?\newline____\_\_\_\_ centimeters by ____\_\_\_\_ centimeters

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Q. The area of a rectangular ink pad is 48square centimeters48\,\text{square centimeters}. The perimeter is 28centimeters28\,\text{centimeters}. What are the dimensions of the ink pad?\newline____\_\_\_\_ centimeters by ____\_\_\_\_ centimeters
  1. Define Variables: Let ll be the length and ww be the width of the rectangular ink pad.\newlineThe area of a rectangle is given by the formula A=l×wA = l \times w.
  2. Area and Perimeter Formulas: The area of the ink pad is given as 4848 square centimeters.\newlineSubstitute 4848 for AA in the area formula.\newline48=l×w48 = l \times w
  3. Equations Setup: The perimeter of a rectangle is given by the formula P=2l+2wP = 2l + 2w.\newlineThe perimeter of the ink pad is given as 2828 centimeters.\newlineSubstitute 2828 for PP in the perimeter formula.\newline28=2l+2w28 = 2l + 2w
  4. Solve Simultaneously: Divide the perimeter equation by 22 to simplify.\newline282=(2l+2w)2\frac{28}{2} = \frac{(2l + 2w)}{2}\newline14=l+w14 = l + w
  5. Rearrange Perimeter Equation: We now have two equations with two variables:\newline48=l×w48 = l \times w (Area equation)\newline14=l+w14 = l + w (Perimeter equation)\newlineWe can solve these equations simultaneously to find the values of ll and ww.
  6. Substitute and Expand: Rearrange the perimeter equation to express one variable in terms of the other. \newlinew=14lw = 14 - l
  7. Quadratic Equation: Substitute w=14lw = 14 - l into the area equation.48=l×(14l)48 = l \times (14 - l)
  8. Factorize: Expand the equation and form a quadratic equation.\newline48=14ll248 = 14l - l^2\newlineMove all terms to one side to set the equation to zero.\newlinel214l+48=0l^2 - 14l + 48 = 0
  9. Solve for Length: Factor the quadratic equation.\newline(l6)(l8)=0(l - 6)(l - 8) = 0
  10. Final Solutions: Solve for ll by setting each factor equal to zero.\newlinel6=0l - 6 = 0 or l8=0l - 8 = 0\newlinel=6l = 6 or l=8l = 8
  11. Final Solutions: Solve for ll by setting each factor equal to zero.l6=0l - 6 = 0 or l8=0l - 8 = 0l=6l = 6 or l=8l = 8If l=6l = 6, then w=14l=146=8w = 14 - l = 14 - 6 = 8.If l=8l = 8, then w=14l=148=6w = 14 - l = 14 - 8 = 6.Both pairs (6,8)(6, 8) and l6=0l - 6 = 000 are valid solutions since a rectangle can have its length and width interchanged.

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