SOLUTION: Sam and Susy had a 40-by-50 foot rectangular yard. Susy cultivated 2/5 of the area of the yard in flowers, which she grew in an even border around the central grass. Every morning

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Question 863747: Sam and Susy had a 40-by-50 foot rectangular yard. Susy cultivated 2/5 of the area of the yard in flowers, which she grew in an even border around the central grass. Every morning Sam walked the dog once around the central grass area inside the flowers. What distance did he cover in his morning stroll with the dog?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
3%2F5 of the rectangular yard is the central grass area.

%2840-2x%29%2850-2x%29=%283%2F5%29%2840%2A50%29 using x as the width of the flower border.
%2840-2x%29%2850-2x%29=3%2A40%2A10
4%2820-x%29%2825-x%29=3%2A40%2A10
%2820-x%29%2825-x%29=300
500-45x%2Bx%5E2=300
x%5E2-45x%2B200=0
discrim, 45^2-4*200=2025-800=1225=49*25
x=%2845-sqrt%2849%2A25%29%29%2F2 because the positive sqrt form will not make sense.
x=%2845-35%29%2F2
x=10%2F2
x=5
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Dimensions of the grass area are 40-2*5 by 50-2*5;
or 30 by 40 square yards.
Perimeter of this grass area is highlight_green%283%2830%2B40%29=highlight%28210%29%29 yards.