To find the area and perimeter of a rectangle you need to know formulas for each: Area of a rectangle=L*W Perimeter of a rectangle=2(L+W) So to find either the area or the perimeter of a rectangle you need to know to things. The value of: The rectangles length The rectangles width To illustrate lets suppose that the length of the rectangle is 2 and the width is 3. Hence: length= 2 width= 3 Area=2*3 Area=6 Perimeter=2(2+3) Perimeter=2(5) Perimeter=10

It is not true that all rectangles of a given area will have the same perimeter. To prove this to yourself, imagine a rectangle measuring 2 x 2, with a perimeter of 8, and area of 4. Then compare to a rectangle measuring 1 x 4, with a perimeter of 10, but with the same area of 4. To solve the problem of your question, you will need to do some algebra, as follows: Perimeter = P = 2 (Length) + 2(Width) = 2L + 2W Area = A = (Length)(Width) = LW Rearrange the equation for perimeter to find W: P = 2L + 2W = 2( L + W) (P/2) = L + W W = (P/2) - L If you substitute this value into the Area equation, A = LW = L( (P/2) - L) = (PL/2)-(L^2) A = -L^2 + (PL/2) Or, rearranging into standard form, L^2 -(P/2)L+A=0 Now you can use the quadratic formula to solve for L.

Remember that for an equation in standard form, aX^2 + bX + c = 0

x = (-b +/- sqrt(b^2 - 4ac)) / 2a

So our equation (where x is L) has a = 1 b = (P/2) c = A

Which gives

L = (-(P/2) +/- sqrt((P/2)^2 - 4A)) / 2

Now, given perimeter and area, you can solve for the dimensions. As an example, P = 15, A = 14

L = (-(15/2) +/- sqrt((15/2)^2 - 4*14)) / 2

L = (-7.5 +/- .5) / 2 = -8 / 2 OR -7 / 2

L = 4 OR 3.5 (You can change the length to a positive number because when measuring length, positive or negative just depends on the direction)

You get 2 answers due to the way the quadratic formula works. However, it makes sense, because you would get two different answers depending on whether the Length or Width of your rectangle was the long dimension. So, choosing L=4, the dimensions of the recangle would be 4 x 3.5 . As a check, A = LW = (4)(3.5) = 14 p = 2L + 2W = 2(4) + 2(3.5) = 8 + 7 = 15

You can write a quadratic equation and solve it. area = L*W perimeter = 2(L+W)

area = L*(perimeter/2 - L) (substitute for W) L^{2} - (perimeter/2)*L + area = 0 (rearrange to standard form) L = (perimeter/2 ±√((perimeter/2)2 - 4(area)))/2 (use the quadratic formula) L = perimeter/4 ± √(perimeter^{2}/16 - area) (simplify)

Please note that there are two possible values for L. If you were to solve for W, you would get the same equation. This means that W is one of the values and L is the other. If you want L > W, then you will choose L = perimeter/4 + √((perimeter/4)^{2} - area) W = perimeter/4 - √((perimeter/4)^{2} - area)

Example Suppose we know the perimeter of a 2x3 plot is 10, and its area is 6. Our formulas give L = 10/4 + √((10/4)2-6) = 2.5 + √(6.25-6) = 2.5+.5 = 3 W = 2.5 - .5 = 2 _____ You will note that L = W = perimeter/4 = √area if the figure is a square.

The length of a room, rectangular in shape, is 8 feet greater than its width. If the perimeter of the room is 64 feet, what are the dimensions of the room?

The perimeter of a rectangle is 170 inches. If the length is 5 more than three times the width what is the dimensions? Is this the formula 2L(5+3w)+2w=170 ?

The perimeter of a rectangle is 80. The length is 5 feet longer than 4 times the width. What are the dimensions of the rectangle? What is the area of the rectangle?