Area and perimeter

Question

Create rectangle with a perimeter that is 4 times as large as the area.

Answer 1

Mark M is correct.  If the measurements are in feet then

the perimeter is in feet and the area is in square feet... and so forth:

 

However: I think below is what you are trying to do...

 

We want P = 4A

 

P = 2L + 2W

A = LW

 

2L + 2W = 4LW

2(L+W) = 4LW

L+W = 2LW

 

If the width and length are both 1 unit then

 

1 unit + 1 unit = 2(1unit)(1unit)

2 units = 2 square units

 

Still,... this is not truly correct because of the difference

between units and square units.

Answer 2

While I agree with Mark M, we can ignore the units for the sake of 'getting an answer' and come up with a numerical answer to this:

the rectangle has a length = x and a width = y, so the perimeter = 2x + 2y

the area is x*y

we're told the perimeter = 4 times the area

so 2x + 2y = 4*xy

we can solve for one variable only, because we have only one equation - though we have two unknowns.  That's ok, because you're only asked to 'create a rectangle' - not find a specific one

 

if y = 2, then 2x + 4 = 4*x*2

2x + 4 = 8x

4 = 6x

x = 4/6 = 2/3

so a rectangle that works, for example, is one with a length of 2/3 and a width of 2.

 

Check this:

the perimeter would be 2/3 + 2/3 + 2 + 2 = 4 + 4/3 = 16/3

the area would be 2/3*2 = 4/3, 4 times the area is 16/3