to find what sides a rectangle must have if its area and perimeter are to be equal?

Answer 1

A = LW
P = 2L + 2W

LW = 2L + 2W
LW - 2L = 2W
L(W - 2) = 2W
L = 2W/(W-2)

So the rectangle can be any size as long as:

L = 2W/(W - 2)

Answer 2

Area and perimeter can be equal.

Area is the number of blocks in the rectangle. Or length times width. For example, if you have four blocks across and four blocks down the number of blocks would be 16. The area would be 16. (a square, not a rectangle)

The perimeter is the length around the outside of the rectangle. If there are 16 marks around the edge, the perimeter is 16. (again, a square, not a rectangle)

Perhaps if you gave us an example of the problem you asking about. Are you given some sides???

I teach 3rd grade, which is a bit more amature than what you are looking for, I am guessing. Maybe this will help or confuse you more. Best of luck!!

Answer 3

I hope this is not a question your teacher asked you--If so, he or she does not understand something about length and area.

Simply put, a length value (like permiter) cannot ever be equal to an area value, because their units are incompatible.

Here's an example of what I mean. Suppose you have a square that is 4 feet on a side. Your first response might be, "Ah! The perimeter is 16 feet, and the area is 16 square feet. So the perimiter and the area are the same!"

But now show that very same square to your friend in France. From his point of view, the perimiter is 4.88 meters, and the area is 1.49 square meters. Now the perimiter does not seem "equal" to the area!

A length may be equal to a length (regardless of units); and an area may be equal to an area (regardless of units). But a length cannot be equal to an area.

Answer 4

lets call the length of the long side x and the length of the short side y

the perimeter of the rectangle is 2*x + 2*y or P=2x+2y

the area is x*y or A=xy,

for the perimeter and area to be equal, you would set P=A or

2x + 2y =xy

this gives you 1 equation for 2 unknowns, so you'll end up with a ratio of x to y
solving for x gives you
x= 2y/(y-2)

so plug in any value for y to get the resulting value for x which will give you dimensions for your rectangle


darn, T was faster, but didn't really explain what he was doing, that first answerer is a retard


True that th units don't match, as area and length are 2 different dimensions, on is in length, the other is length^2.

Answer 5

An area and a length can never be equivalent.

EDIT:
let me elaborate if I have a rectangle 3 cm X 6 cm the the perimeter is 18 cm and the area is 18 sq cm.
Now in inches the perimeter is 7.09 inches and the area is 2.79 sq inches.. so go figure

Answer 6

First responder is an idiot, a square of 1foot per side has an area of 1.

Answer 7

"...if its area and perimeter are to be equal?"
If the equations for area & perimeter are set equal?
lw = 2l + 2w
Ex.Solve for l.
lw - 2l = 2w
l(w - 2) = 2w
l = 2w / (w - 2).

Answer 8

this must mean if the sides are a and b then

ab = 2(a + b)
ab = 2a + 2b
ab - 2a = 2b
a(b- 2) = 2b
a = 2b/(b-2)

plug in a value for "A" and get value for "b"

Answer 9

Let L = length and W = width.
Area = LW
Perimeter = 2L + 2W

Equation:
LW = 2L + 2W
L = (2L + 2W) / W
L = 2L/W + 2