SOLUTION: I need to answer this question using quadratics and I am not sure where to start. Ed needs to enclose a rectangular section of his yard. The area is 35 sq ft and the perimeter is 2
Question 166189: I need to answer this question using quadratics and I am not sure where to start. Ed needs to enclose a rectangular section of his yard. The area is 35 sq ft and the perimeter is 27 ft. Find the length and the width of the section. Found 2 solutions by vleith, Mathtut:Answer by vleith(2983) (Show Source):
You can put this solution on YOUR website! Perimeter is the length 'around' the rectangular area.
Area is the area of the enclosed rectangle
Now use the quadratic equation to solve
You can put this solution on YOUR website! since A=lw where l is length and w is the width
and P=2(l+w)
we plug in the known values
35=lw
27=2(l+w)--->27/2=l+w---->l=27/2-w---->l=(27-2w)/2
place value of l into 1st equation----->35=((27-2w)/2)(w)-----> factoring (2w-7)(w-10) so w=7/2,10
when w=7/2 l=10 when w=10 l=7/2