1. Polynomial story problem.

Set up a polynomial equation and solve it in order to answer the question:

A rectangular parcel of land is 50ft. wide. The length of a diagonal between opposite corners is 10ft. more than the length of the parcel. What is the parcels length?

Any help is very much appreciated.

2. Originally Posted by zagsfan20
Set up a polynomial equation and solve it in order to answer the question:

A rectangular parcel of land is 50ft. wide. The length of a diagonal between opposite corners is 10ft. more than the length of the parcel. What is the parcels length?...
Hello,

a rectangle consists of two congruent right triangles. The diagonal ofthe rectangle is the hypotenuse of the right triangle.

You can use Pythagorian rule:

Let l be the length of the rectangle
let d be the diagonal of the rectangle: d = l +10

Then you get the equation:

$\displaystyle 50^2+l^2=(l+10)^2 \ \Longleftrightarrow \ 2500+l^2=l^2+20l+100$ Solve for l.

I've got l = 120 ft.

EB

3. Hello, zagsfan20!

Did you make a sketch?

Set up a polynomial equation and solve it in order to answer the question.

A rectangular parcel of land is 50 ft wide.
The length of a diagonal is 10 ft more than the length.
What is the parcel's length?
Code:
      * - - - - - - - - - - - - - *
|                       *   |
|                   *       |
|          L+10 *           |
|           *               | 50
|       *                   |
|   *                       |
* - - - - - - - - - - - - - *
L
Then, as earboth pointed out: .$\displaystyle (L+10)^2\:=\:L^2 + 50^2$

Go for it!