SOLUTION: there is a rectangular section of his yard. the perimeter of the section is 27 feet and the are is 35 square feet find the length and width of the section. p=2(L+W) if the perimete

Question 1024232: there is a rectangular section of his yard. the perimeter of the section is 27 feet and the are is 35 square feet find the length and width of the section. p=2(L+W) if the perimeter is 27 feet solve for W. The equation for area is A=LW, substitute the expression you found for W in part 1 into the area equation. Then, if the area is equal to 35 square feet write the equation in standard form with a,b, and c as whole numbers. Then use the quadratic formula to solve for L. PLEASE HELP!
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
(small spelling correction given here)......
...rectangular section of his yard. the perimeter of the section is 27 feet and the area is 35 square feet .

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FIND DIMENSIONS OF RECTANGLE USING GIVEN PERIMETER & AREA
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

there is a rectangular section of his yard. the perimeter of the section is 27 feet and the are is 35 square feet find the length and width of the section. p=2(L+W) if the perimeter is 27 feet solve for W. The equation for area is A=LW, substitute the expression you found for W in part 1 into the area equation. Then, if the area is equal to 35 square feet write the equation in standard form with a,b, and c as whole numbers. Then use the quadratic formula to solve for L. PLEASE HELP!

P = 2(L + W)
27 = 2(L + W)
2(13.5) = 2(L + W)
13.5 = L + W

A = LW
A = L(13.5 - L) -------- Substituting 13.5 - L for W
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You are told to use the quadratic equation formula to solve for L, so I'll leave that task up to you to complete this problem.
If it were left to me though, I would multiply the equation by 2 to get: , and then solve by factoring to
get dimensions:
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