SOLUTION: Tress are planted according to the "square" system, 30 ft. apart each way. A different style, "hexagonal" utilizes more space available (all but 10% of ground is used). Trees are
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Question 1051556: Tress are planted according to the "square" system, 30 ft. apart each way. A different style, "hexagonal" utilizes more space available (all but 10% of ground is used). Trees are planted by this system 30 ft. apart. Determine number of trees per acre.
Non-homework.
Hexagon has 6 equal sides.
Equilateral triangle has 3 equal sides.
30 * 3 = 90.
39204(10% deducted from 43560) / 90 = 435.6 = 436 trees.
Not sure.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Tress are planted according to the "square" system, 30 ft. apart each way. A different style, "hexagonal" utilizes more space available (all but 10% of ground is used). Trees are planted by this system 30 ft. apart. Determine number of trees per acre.
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The trees would be 30 ft apart in horizontal rows.
The vertical row spacing is sqrt(675) ft =~ 25.98 feet
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Using a rectangular plot 210 by 207.43 feet:
--> 207.43 = 7.98 = 8 trees vertically
210/30 = 7 trees horizontally.
-- 56 trees/acre
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Using a square acre 208.71 ft sides:
208.71/25.98 = 8.033 trees/row
208.71/30 =6.96 trees/row
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If it's 1 acre, you get 48 or 49 per acre.
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The area of each hexagon is ~ 2338 sq ft
Area = ns^2*cot(180/n)/4 where n = # of sides and s = side length.
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43560/2338 =~ 18.63 hexagons
Each hexagon is 7 trees, but 6 of them are part of 3 different hexagons.
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Consider each tree a circle with a radius of 15 feet.
The area of each circle is ~ 706.86 sq feet
43560/706.86 =~ 61 trees
It's 61 trees/acre if all the area is used.
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Using a hexagonal acre:
The trees would arranged with 1 in the center, 6 surrounding it, then 12 more, then 18 more, then 24 more, etc
# of hexagons = 1, 7, 19, 37, 61, etc.
# of hexagons/acre =~ 18.63
The # of trees for 19 hexagons = 50 (I counted them).
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Depending on the shape of the acre used for measurement, 48 to 50 trees/acre.
Packing & stacking problems are often complex, even in 2 dimensions.
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