SOLUTION: A lawn is 90 feet long and 40 feet wide. By cutting along the perimeter, how wide is this border if 1/3 of the grass is cut? More info: provided solution is 5 feet. Student s

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: A lawn is 90 feet long and 40 feet wide. By cutting along the perimeter, how wide is this border if 1/3 of the grass is cut? More info: provided solution is 5 feet. Student s      Log On

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Question 579237: A lawn is 90 feet long and 40 feet wide. By cutting along the perimeter, how wide is this border if 1/3 of the grass is cut?

More info: provided solution is 5 feet. Student says answer must be derived by using quadratic formula.
X=width of the border
2(90)(x)+2(40)(x)=(90 times 40)/3
Which then does not result in a constant for a to be used in the quadratic formula. B is 260 and c is negative 1200. Since the constant a is zero, i get a divide by zero using the quadratic formula which means to me the answer is infinite but the provided answer is 5. I have made an error but i can't find it. Please advise.

Found 2 solutions by mananth, josmiceli:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
Area of garden = 90 * 40 ft = 3600 sq. ft
1/3 of this area is cut = 1200 sq. ft
Let the uniform width be x
so the reduced length = 90-2x
reduced width = 40-2x
area of this portion is ( 90-2x)((40-2x)
Total area - reduced area = area of the grass cut
3600-(90-2x)(40-2x)= 1200
(90-2x)(40-2x)= 3600-1200
90(40-2x)-2x(40-2x) = 2400
3600-180x-80x+4x^2=2400
4x^2-260x+1200=0
/4
x^2-65x+300=0
x^2-60x-5x+300=0
x(x-60)-5(x-60)=0
(x-60)(x-5)=0
x= 60 not possible
x= 5 ft the width
m.ananth@hotmail.ca



Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The area of the grass is +90%2A40+=+3600+ ft2
1/3 of the area is +1200+ ft2
If the border is of uniform width, and this width
is +x+, then the total area minus the border is
+%2840+-+2x%29%2A%2890+-+2x%29+=+3600+-+1200+
+3600+-+180x+-+80x+%2B+4x%5E2+=+2400+
+4x%5E2+-+260x+%2B+1200+=+0+
+x%5E2+-+65x+%2B+300+=+0+
Using quadratic formula:
+x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+-65+
+c+=+300+
+x+=+%28-%28-65%29+%2B-+sqrt%28+%28-65%29%5E2+-+4%2A1%2A300+%29+%29+%2F+%282%2A1%29+
+x+=+%28+65+%2B-+sqrt%28++4225+-4%2A1%2A300+%29%29%2F%282%2A1%29+
+x+=+%28+65+%2B-+sqrt%28++4225+-+1200+%29+%29+%2F+2+
+x+=+%28+65+%2B-+sqrt%28+3025+%29+%29+%2F+2+
+x+=+%28+65+%2B-+55+%29+%2F+2+
+x+=+10%2F2+
+x+=+5+
and, with negative square root,
+x+=+120%2F2+
+x+=+60+ ( not possible )
The border is 5 ft wide
check:
+%2840+-+2x%29%2A%2890+-+2x%29+=+3600+-+1200+
+%2840+-+2%2A5%29%2A%2890+-+2%2A5%29+=+3600+-+1200+
+%28+40+-+10+%29%2A%28+90+-+10+%29+=+3600+-+1200+
+30%2A80+=+2400+
+2400+=+2400+
OK