A field hockey tournament is to be held in Yellowknife, NWT. According to regulations, the field is 25 yards by 100 yards. For officials, support groups and team members, a uniform rectangular strip of manicured lawn is to be installed by a local business group. If the area of the manicured lawn is twice the surface area of the playing field, how wide will this strip be to the nearest tenth of a meter?

Question

A field hockey tournament is to be held in Yellowknife, NWT. According to regulations, the field is 25 yards by 100 yards. For officials, support groups and team members, a uniform rectangular strip of manicured lawn is to be installed by a local business group. If the area of the manicured lawn is twice the surface area of the playing field, how wide will this strip be to the nearest tenth of a meter?
Need help for how to solve this equation.

Henry · Answer

A = 2 * 25 * 100 = 5000 Sq. Yds. = Area of the lawn, A = L * W = 5,000, 100W = 5,000, W = 50 Yds.

bobpursley · Answer

well, the dimensions of each end is (25+2w)w and the side (100w). the total lawn area
A=2(25+2w)w+2(100w)=50w+4w^2+200w=4w^2+250w=2*2500
4w^2+250w-5000=0
w^2+62.5w-1250=0
solve the equation.
check my thinking

d

Damon · Answer

well, (100+2w)(25+2w) = 3 (100*25) 2500 + 250 w + 4 w^2 = 7500 w^2 + 62.5 w - 1250 = 0 agree with bobpursley