# Dimensions of Rectangle

• Dec 6th 2008, 07:16 AM
magentarita
Dimensions of Rectangle
A rope 100 feet long is stretched around four posts set at the corners of a rectangle whose area is 576 square feet. Find the dimensions of the rectangle.
• Dec 6th 2008, 07:24 AM
janvdl
Quote:

Originally Posted by magentarita
A rope 100 feet long is stretched around four posts set at the corners of a rectangle whose area is 576 square feet. Find the dimensions of the rectangle.

Since the rope is 100 feet long, then the perimeter of this rectangle must be 100 feet long too.

So we can say:

\$\displaystyle P = 100 = 2L + 2W\$

And simplify that to:

\$\displaystyle 50 = L + W\$

And let's rearrange that to:

\$\displaystyle L = 50 - W\$

---

Now let's go to the area (This is going to end up as a simul. equation)

\$\displaystyle A = 576 = LW\$

Now let's substitute that value we have for the length

\$\displaystyle 576 = (50 - W)(W)\$

\$\displaystyle 576 = 50W - W^2\$

\$\displaystyle W^2 - 50W + 576 = 0\$

Solve for W, and subsequently L.
• Dec 7th 2008, 09:03 AM
magentarita
ok.....
Quote:

Originally Posted by janvdl
Since the rope is 100 feet long, then the perimeter of this rectangle must be 100 feet long too.

So we can say:

\$\displaystyle P = 100 = 2L + 2W\$

And simplify that to:

\$\displaystyle 50 = L + W\$

And let's rearrange that to:

\$\displaystyle L = 50 - W\$

---

Now let's go to the area (This is going to end up as a simul. equation)

\$\displaystyle A = 576 = LW\$

Now let's substitute that value we have for the length

\$\displaystyle 576 = (50 - W)(W)\$

\$\displaystyle 576 = 50W - W^2\$

\$\displaystyle W^2 - 50W + 576 = 0\$

Solve for W, and subsequently L.

Thank you. I had a hard time setting up the right equation.