1. P = 2(L + W) The preceding formula gives the perimeter P of a rectangle with length L and width W. An outdoor patio is in the shape of a rectangle and has a perimeter of 48 feet. The width of the patio is 4 feet less than the length of the patio. What is the width, in feet, of the patio?
Answer: A
The width of the patio is 10 feet.
To find the width of the patio, we use the formula for the perimeter of a rectangle, P = 2(L + W), and the information given. Given that the perimeter is 48 feet and the width is 4 feet less than the length, we can set up the equations and solve for the width to find that it is indeed 10 feet.
A) 10
This option is correct. If we let the length be L and the width be W, we know that W = L - 4. Substituting into the perimeter formula gives us 48 = 2(L + (L - 4)), which simplifies to 48 = 2(2L - 4), leading to 48 = 4L - 8. Solving this equation results in L = 14, and therefore W = 14 - 4 = 10.
B) 12
This option is incorrect. If the width were 12 feet, then according to the relationship W = L - 4, the length would be 12 + 4 = 16 feet. Substituting these values into the perimeter formula yields P = 2(16 + 12) = 2(28) = 56 feet, which does not match the given perimeter of 48 feet.
C) 20
This option is incorrect. If the width were 20 feet, then the length would be 20 + 4 = 24 feet. Substituting these values into the perimeter formula gives P = 2(24 + 20) = 2(44) = 88 feet, which is far greater than the required perimeter of 48 feet.
D) 24
This option is incorrect. If the width were 24 feet, the length would be 24 + 4 = 28 feet. Using these dimensions in the perimeter formula results in P = 2(28 + 24) = 2(52) = 104 feet, which is again not equal to the specified perimeter of 48 feet.
Conclusion
The correct answer, 10 feet, is derived from accurate calculations based on the perimeter formula and the relationship between length and width. All other options do not satisfy the conditions set by the problem, thereby confirming that they are incorrect. The calculations clearly demonstrate that the only viable solution is a width of 10 feet.