SOLUTION: A park has a circular lake 240' in diameter. A walkway 6' wide is around the lake. Determine number of sq. yds. in the walk. A = area. 240' = 80 yds., 6' = 2 yds. A =

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: A park has a circular lake 240' in diameter. A walkway 6' wide is around the lake. Determine number of sq. yds. in the walk. A = area. 240' = 80 yds., 6' = 2 yds. A =      Log On

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Question 1084974: A park has a circular lake 240' in diameter. A walkway 6' wide is around the lake. Determine number of sq. yds. in the walk.
A = area.
240' = 80 yds., 6' = 2 yds.
A = pi(40 + 2)(40 - 2)
A = pi(42)(38)
A = pi(1596)
A = 5013.9936 = 5014 sq. yds.
Area of circle: pi * r^2
Area of lake: pi * 40 * 40 = 5026.56 = 5027 sq. yds.
Area of lake and walk: pi * 42 * 42 = 5541.7824 = 5542 sq. yds.
Area of walk: 5542 - 5027 = 515 sq. yds.
Where am I incorrect. Non-homework. Thanks.
Found 2 solutions by Fombitz, MathLover1:
Answer by Fombitz(32388) About Me  (Show Source):
Answer by MathLover1(20849) About Me  (Show Source):
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A park has a circular lake 240' in diameter.->
240in+=+6.67yd
then, radius is 6.67yd%2F2=3.33yd
area of the lake is
%283.33yd%29%5E2%2Api=11.1%2Api%2Ayd%5E2
A walkway 6inwide is around the lake;
6in+=+0.17yd
so, radius of the lake and walkway is 3.33yd%2B0.17yd=3.5yd
area of the lake and walkway:
%283.5yd%29%5E2%2Api=12.25%2Api%2Ayd%5E2
area of the walkway alone is:
12.25%2Api%2Ayd%5E2-11.1%2Api%2Ayd%5E2=1.15%2Api%2Ayd%5E2=1.15%2A3.14%2Ayd%5E2=3.611yd%5E2