Is this impossible?

Question

Is it impossible to make a figure with a perimeter of 30 and an area of 9?

If it is impossible, please explain why.

Answer 1

Of course it is not impossible.
Make a rectangle with side of 0.63 and 14.37.
This will give an area of 9.053

Answer 2

If it is a rectangle:
P = 2*(l + w) = 30
l+w = 15
l = 15-w

A = l * w = 9
(15-w)*w = 9
15w - w^2 = 9
w^2 -15w +9 = 0
by quadratic formula
w = 15/2 +/- sqrt(225-36)
w = 7.5 +/- sqrt(189)
w ~= 21.25

However: if w ~= 21.25 and l = 15-w, then l < 0.
Therefore, this figure cannot be constructed as a rectangle. It may be possible to construct a figure with perimeter 30 and area 9 as some other polygon, but i don't think it would be...

Answer 3

No

Rectangle possibility

Side 1 .62
Side 2 14.38

These are estimates from a quadratic

2*.62 + 2 * 14.38 = 30 Perimeter
.62 * 14.38 = 9 AREA

Answer 4

This would give the equations as follows where x is the height and y is the width.

2x + 2y = 30
xy = 9

I believe that their is a solution to these. The perimeter of a rectangle can be make arbitrarily large with the same area by making the rectangle as long and skinny as you need it to be.

Answer 5

No, this is possible.
For a rectangle,
Perimeter= 2*(length + width)=30
Area=length*width=9
2(l+w)=30 => l+w=15
l=15-w
(15-w)*w=9
15*w-w^2=9
w^2-15w+9=0
w=(15 +/- sqrt(225-4*9))/2
w=(15+sqrt(189))/2 ~= 14.3738635
l=(15-sqrt(189))/2 ~= 0.626136458

Check: 2*(0.626136458+14.3738635) = 30
0.626136458*14.3738635= 9

Answer 6

ya it is impossible cuz the perimeter cannot be 30 and the area be 9 come on its comomn sense

Answer 7

no

let assume the figure is a rectan. with lengh l and width w

perimeter = (l + w)*2=30 and area = w*l = 9

i think u are able to find l and w and there is a solution.

Answer 8

It is possible because 30 and 9 could be anything, they haven't been defined.

Answer 9

Yes

A rectangle of 14.3735 x 0.6261

Answer 10

yes