SOLUTION: For the given area, find the length and width of the rectangle with the least perimeter. Use whole numbers only.
30 m2
40 yd2
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-> SOLUTION: For the given area, find the length and width of the rectangle with the least perimeter. Use whole numbers only.
30 m2
40 yd2
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Question 194543This question is from textbook Harcourt Math
: For the given area, find the length and width of the rectangle with the least perimeter. Use whole numbers only.
30 m2
40 yd2 This question is from textbook Harcourt Math
You can put this solution on YOUR website! For the given area, find the length and width of the rectangle with the least perimeter. Use whole numbers only.
30 m2
40 yd2
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A square has the least perimeter for a rectangle. (A proof is available is needed.)
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For squares, P = 4*s
30 sq meters --> sides of sqrt(30) meters
l = w = sqrt(30) = ~ 5.477 meters
Using integers, --> 5 by 6.
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40 sq yds --> sqrt(40)
l = w = sqrt(40) = ~ 6.325 yards
Using integers --> 5 by 8 yards
This is a silly problem.
