# Polynomial story problem.

• Jan 12th 2007, 12:00 AM
zagsfan20
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.
• Jan 12th 2007, 01:27 AM
earboth
Quote:

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
• Jan 12th 2007, 05:30 AM
Soroban
Hello, zagsfan20!

Did you make a sketch?

Quote:

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!