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The area of a rectangular garage is 30square meters30 \, \text{square meters}. The perimeter is 22meters22 \, \text{meters}. What are the dimensions of the garage?\newline____\_\_\_\_ meters by ____\_\_\_\_ meters

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Q. The area of a rectangular garage is 30square meters30 \, \text{square meters}. The perimeter is 22meters22 \, \text{meters}. What are the dimensions of the garage?\newline____\_\_\_\_ meters by ____\_\_\_\_ meters
  1. Define Variables: Let ll be the length and ww be the width of the rectangular garage.\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 garage is given as 3030 square meters.\newlineSubstitute 3030 for AA in the formula A=l×wA = l \times w.\newline30=l×w30 = l \times w
  3. Substitute Values: The perimeter of a rectangle is given by the formula P=2l+2wP = 2l + 2w. The perimeter of the garage is given as 2222 meters. Substitute 2222 for PP in the formula P=2l+2wP = 2l + 2w. 22=2l+2w22 = 2l + 2w
  4. Simplify Perimeter Equation: Divide the perimeter equation by 22 to simplify.\newline222=2l+2w2\frac{22}{2} = \frac{2l + 2w}{2}\newline11=l+w11 = l + w
  5. Solve System of Equations: We now have two equations with two variables:\newline30=l×w30 = l \times w (Area equation)\newline11=l+w11 = l + w (Perimeter equation)\newlineWe can solve this system of equations to find the values of ll and ww.
  6. Express Width in Terms of Length: From the perimeter equation, express ww in terms of ll.w=11lw = 11 - l
  7. Substitute Width into Area Equation: Substitute w=11lw = 11 - l into the area equation.30=l×(11l)30 = l \times (11 - l)
  8. Distribute and Rearrange Equation: Distribute ll across the parentheses. 30=11ll230 = 11l - l^2
  9. Factor Quadratic Equation: Rearrange the equation into a standard quadratic form.\newline0=l211l+300 = l^2 - 11l + 30
  10. Solve for Length: Factor the quadratic equation.\newline(l5)(l6)=0(l - 5)(l - 6) = 0
  11. Find Width: Solve for ll by setting each factor equal to zero.l5=0l - 5 = 0 or l6=0l - 6 = 0l=5l = 5 or l=6l = 6
  12. Final Valid Solutions: If l=5l = 5, then w=11l=115=6w = 11 - l = 11 - 5 = 6. If l=6l = 6, then w=11l=116=5w = 11 - l = 11 - 6 = 5. Both pairs (5,6)(5, 6) and (6,5)(6, 5) are valid solutions since a rectangle can have its length and width interchanged.

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