Derive equation of rectangle

• Sep 14th 2010, 05:03 PM
webguy
Derive equation of rectangle
Hi,

I have a triangle whose perimeter is 34cm. The diagonal is 13cm, and the width is $x$cm. Derive the equation $x^2-17x+60=0$.

I attempted this, and this is what I did:
Using Pythagoras' theorem I knew that $13 = x^2 + h^2$ where $h$ is the height of the rectangle. Thus, the unknown must be $h^2 = 13-x^2$ which becomes $h = \sqrt{13-x^2}$, therefore the following should be true: $2x+2(\sqrt{13-x^2})=34$. However as you can see it doesn't match the derivation given at the start. How wrong was my answer? Where do I go from here?

I would like to point out that this isn't homework.

Thanks for any help (Happy)
• Sep 14th 2010, 05:15 PM
webguy
Since seeing it on screen I have managed to progress:

$2(\sqrt{13-x^2})=34-2x$
$\sqrt{13-x^2}=17-x$
$13-x^2=(17-x)^2$
$13-x^2=289-34x+x^2$
$276-34x+2x^2=0$
$138-17x+x^2=0$

However this still doesn't match the provided derivation...
• Sep 14th 2010, 05:48 PM
Quote:

Originally Posted by webguy
Since seeing it on screen I have managed to progress:

$2(\sqrt{13-x^2})=34-2x$
$\sqrt{13-x^2}=17-x$
$13-x^2=(17-x)^2$
$13-x^2=289-34x+x^2$
$276-34x+2x^2=0$
$138-17x+x^2=0$

However this still doesn't match the provided derivation...

You have two major typos!!

According to your calculations, the shape ought to be a rectangle,
so that the perimeter is 2x+2h.

Then you are incorrectly applying Pythagoras' theorem on top of that....

$x^2+h^2=13^2$

The square on the hypotenuse is the sum of the squares of the perpendicular sides.

Hence, replace your 13 in the calculations with 169 and see how you get on..

EDIT: my apologies....you do have "rectangle" in your title!
• Sep 14th 2010, 07:22 PM
Soroban
Hello, webguy

Quote:

I have a rectangle whose perimeter is 34 cm.
The diagonal is 13 cm, and the width is $x$ cm.

Derive the equation: $x^2-17x+60\:=\:0$

Code:

    * - - - - - - - - - - - *     |                  *  |     |        13    *      |     |          *          | x     |      *              |     |  *                  |     * . - - - - - - - - - - *                 L

The length of the rectangle is $\,L.$
The width of the rectangle is $\,x$.

The perimeter is 34: . $2L + 2x \:=\:34 \quad\Rightarrow\quad L \:=\:17 - x$ .[1]

Pythagorus says: . $x^2 + L^2 \:=\:13^2$

Substitute [1]: . $x^2 + (17-x)^2 \:=\:169$

. . . . . . . $x^2 + 289 - 34x + x^2 \:=\:169$

. . . . . . . . . . $2x^2 - 34x + 120 \;=\;0$

. . . . . . . . . . . $x^2 - 17x - 60 \:=\:0$
• Sep 15th 2010, 02:07 AM
webguy
Quote:

Originally Posted by webguy
I have a triangle whose perimeter is 34cm.

Oops! Sorry, that should have said rectangle.

I don't understand why it's $13^2=x^2+h^2$. Pythagoras' theorem states that the hypotenuse is $c^2=b^2-a^2$, and I know the hypotenuse; it's 13. So shouldn't it be $13=b^2-a^2$?
• Sep 15th 2010, 02:55 AM
RHandford
Hi

Sory to be so thick but where does the 34x come from in the second line after substitute?

• Sep 15th 2010, 03:20 AM
webguy
From expanding $(17-x)^2$ you get $289-17x-17x+x^2$.

----------

I don't understand why it's http://www.mathhelpforum.com/math-he...f5d954bd17.png. Pythagoras' theorem states that the hypotenuse is http://www.mathhelpforum.com/math-he...49fef8a326.png, and I know the hypotenuse; it's 13. So shouldn't it be http://www.mathhelpforum.com/math-he...ee8c6d11fd.png?
• Sep 15th 2010, 03:24 AM
Educated
The pythagorus theorum states:

$a^2 + b^2 = c^2$

a and b are the lengths of the triangle
c is the hypotenuse.

You have the pythagorus theorum wrong. The equation you have is to find the other length whilst knowing the hypotenuse and a length.

Wikipedia: Pythagoras theorum
• Sep 15th 2010, 03:26 AM
webguy
Oh wow I am completely stupid for making such a mistake! I will return to my cave...
• Sep 15th 2010, 03:52 AM
webguy
Actually I still don't understand why it's why it's $13^2=x^2+h^2$ and not $13=x^2+h^2$.
• Sep 15th 2010, 04:22 AM
Quote:

Originally Posted by webguy
Actually I still don't understand why it's why it's $13^2=x^2+h^2$ and not $13=x^2+h^2$.

I could draw some diagrams to show you, though Educated's link contains most of them,
but if you use a ruler and draw two perpendicular sides 3 centimeters long and 4 centimeters long.
Then draw the hypotenuse and measure it.
It should be 5 centimeters long if the other two sides are perpendicular.

Notice then that

$4^2+3^2=16+9=25=5^2$

It works the same way for all other dimensions if the sides are perpendicular.
• Sep 15th 2010, 12:30 PM
webguy
I understand Pythagoras' proof, but I don't understand why it is $13^2$ and not $13$. I know the hypotenuse is 13cm, so why do I square it instead of replacing $c^2$?
• Sep 15th 2010, 12:33 PM
undefined
Quote:

Originally Posted by webguy
I understand Pythagoras' proof, but I don't understand why it is $13^2$ and not $13$.

The equation is $\,a^2+b^2=c^2$.

13 takes the place of c, if that helps.
• Sep 15th 2010, 01:24 PM
Quote:

Originally Posted by webguy
I understand Pythagoras' proof, but I don't understand why it is $13^2$ and not $13$. I know the hypotenuse is 13cm, so why do I square it instead of replacing $c^2$?

I think I see what you mean....

The side lengths of the triangle are x, (17-x) and 13.

These are your $a,\; b,\; c.$

You have to square all of those to use Pythagoras' theorem.
The lengths of the perpendicular sides themselves do not sum to give the hypotenuse length,
but the squares of the perpendicular sides do sum to give the square of the hypotenuse.

$a^2+b^2=c^2\Rightarrow\ x^2+(17-x)^2=13^2$

$x^2+289-34x+x^2=169$

$2x^2-34x+289-169=0$

$2x^2-34x+120=0$
• Sep 15th 2010, 03:40 PM
webguy
Thank you for all the help!