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A rectangular rug has an area of 21square feet21 \, \text{square feet}. Its perimeter is 20feet20 \, \text{feet}. What are the dimensions of the rug?\newline____\_\_\_\_ feet by ____\_\_\_\_ feet

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Q. A rectangular rug has an area of 21square feet21 \, \text{square feet}. Its perimeter is 20feet20 \, \text{feet}. What are the dimensions of the rug?\newline____\_\_\_\_ feet by ____\_\_\_\_ feet
  1. Define Rug Area: Let's denote the length of the rug as ll feet and the width as ww feet. The area of the rug is given by the product of its length and width.\newlineWhich equation represents the area of a rectangle.\newlineArea = l×wl \times w
  2. Calculate Area: The area of the rectangular rug is given as 2121 square feet.\newlineWhich equation represents the area of this rectangular rug.\newlineSubstitute 2121 for Area in Area=l×w\text{Area} = l \times w.\newline21=l×w21 = l \times w
  3. Define Perimeter: The perimeter of a rectangle is given by the sum of twice its length and twice its width.\newlineWhich equation represents the perimeter of a rectangle.\newlinePerimeter = 2l+2w2l + 2w
  4. Calculate Perimeter: The perimeter of the rug is given as 2020 feet.\newlineWhich equation represents the perimeter of this rectangular rug.\newlineSubstitute 2020 for Perimeter in Perimeter=2l+2w\text{Perimeter} = 2l + 2w.\newline20=2l+2w20 = 2l + 2w
  5. Solve System of Equations: We now have a system of two equations with two variables:\newline21=l×w21 = l \times w (11)\newline20=2l+2w20 = 2l + 2w (22)\newlineWe can solve this system of equations to find the values of ll and ww.
  6. Simplify Perimeter Equation: Let's simplify equation (22) by dividing all terms by 22 to make the calculations easier.\newline202=(2l+2w)2\frac{20}{2} = \frac{(2l + 2w)}{2}\newline10=l+w10 = l + w\newlineNow we have a simpler equation:\newline10=l+w10 = l + w (33)
  7. Express Width in Terms of Length: We can express ww from equation (33) in terms of ll.w=10lw = 10 - lNow we can substitute this expression for ww into equation (11).
  8. Substitute Width into Area Equation: Substitute w=10lw = 10 - l into 21=l×w21 = l \times w.21=l×(10l)21 = l \times (10 - l)This gives us a quadratic equation:21=10ll221 = 10l - l^2
  9. Rearrange Quadratic Equation: Rearrange the quadratic equation to standard form.\newline0=l210l+210 = l^2 - 10l + 21
  10. Factor Quadratic Equation: Factor the quadratic equation. \newline(l7)(l3)=0(l - 7)(l - 3) = 0
  11. Solve for Length: Set each factor equal to zero and solve for ll.l7=0l - 7 = 0 or l3=0l - 3 = 0This gives us two possible solutions for ll:l=7l = 7 or l=3l = 3
  12. Find Possible Dimensions: If l=7l = 7, then from equation (33) w=10lw = 10 - l, we get w=107=3w = 10 - 7 = 3. If l=3l = 3, then w=103=7w = 10 - 3 = 7. So the dimensions of the rug can be either 77 feet by 33 feet or 33 feet by 77 feet.

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