Asked by Joey
Is it possible to have a figure an area of 9 square units and a perimeter of 30? If not, why? (PS I don't think it is possible.)
Answers
Answered by
bobpursley
LW=9
2L+2W=30
that is for a rectangle.
2*L/9+2L=30
20L=270
L=13.5, W=13.5/9=you do it.
2L+2W=30
that is for a rectangle.
2*L/9+2L=30
20L=270
L=13.5, W=13.5/9=you do it.
Answered by
Steve
sure maximum area for a given perimeter is a square.
A square with perimeter 30 has area 7.5^2 = 56.25
So, since 9 < 56.25, we can find such a rectangle.
If the width is x, then the length is 15-x, so
x(15-x) = 9
x^2 - 15x - 9 = 0
x = 3/2 (5 - √21) = 0.62614
the rectangle has area
= 9/4 (5-√21)(5+√21)
= 9/4 (25-21)
= 9/4 * 4
= 9
A square with perimeter 30 has area 7.5^2 = 56.25
So, since 9 < 56.25, we can find such a rectangle.
If the width is x, then the length is 15-x, so
x(15-x) = 9
x^2 - 15x - 9 = 0
x = 3/2 (5 - √21) = 0.62614
the rectangle has area
= 9/4 (5-√21)(5+√21)
= 9/4 (25-21)
= 9/4 * 4
= 9
Answered by
Molly
LW=9
2L+2W=30
that is for a rectangle.
2*L/9+2L=30
20L=270
L=13.5, W=13.5/9=you do it.
2L+2W=30
that is for a rectangle.
2*L/9+2L=30
20L=270
L=13.5, W=13.5/9=you do it.
Answered by
Steve
W=9/L, not W=L/9
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