if the perimeter is 4 and the area is 8, what is the width?
3 answers
i think its two?
a 1x1 square has perimeter 4 and area 1
a square has the maximum area for a given perimeter.
So, this scenario is not possible.
Now, if the perimeter is 8 and the area is 4, then it's a 2x2 square.
a square has the maximum area for a given perimeter.
So, this scenario is not possible.
Now, if the perimeter is 8 and the area is 4, then it's a 2x2 square.
The perimeter of what???
Assuming you have a rectangle.
let the length be x and the width be y
Area = xy = 8
then y = 8/x
perimeter = 2x + 2y = 4
x + y = 2
x + 8/x = 2
x^2 + 8 = 2x
x^2 - 2x + 8 = 0
x = (2 ± √-28)/2 , which is not a real number
So the problem has no solution if we assume we have a rectangle
When you say, "I think it's two" , what made you say that?
(if your width is 2, then your length would be 4 to get 2*4=8
then the perimeter would have to be 2(2+4) = 12, not the given 4 )
Assuming you have a rectangle.
let the length be x and the width be y
Area = xy = 8
then y = 8/x
perimeter = 2x + 2y = 4
x + y = 2
x + 8/x = 2
x^2 + 8 = 2x
x^2 - 2x + 8 = 0
x = (2 ± √-28)/2 , which is not a real number
So the problem has no solution if we assume we have a rectangle
When you say, "I think it's two" , what made you say that?
(if your width is 2, then your length would be 4 to get 2*4=8
then the perimeter would have to be 2(2+4) = 12, not the given 4 )