Polygon Sides

magentarita · #1 $Old$ January 7th, 2009, 09:38 PM

A polygon with sides: 154-x, 50-2x, x+120, and x^2 + 7x.
Solve for x and give each angle, if necessary.

vincisonfire · #2 $Old$ January 9th, 2009, 10:59 AM

Actually it is not a polygon since it has a curve in it.
Doing a sketch, you should easily find the angle and intersect of the lines.
$*50-2x = x +120$
$-70 = 3x$
$-\frac{70}{3} = x$
This is an example, it is NOT one of the vertexes.
For the parabola, I would use the derivative to find the slope at the intersect.
If needed ask for further explanations.

masters · #3 $Old$ January 9th, 2009, 11:30 AM

Quote:

Originally Posted by magentarita $View Post$

A polygon with sides: 154-x, 50-2x, x+120, and x^2 + 7x.
Solve for x and give each angle, if necessary.

Hello magentarita,

If I were a betting person, I would bet that you meant to say angles instead of sides, because:

$154-x + 50-2x + x+120 + x^2+7x = 360$

since the sum of the angles of a quadrilateral = 360 degrees.

Now, this all boils down to:

$x^2+5x-36=0$

$(x+9)(x-4)=0$

$x=-9 \ \ or \ \ x=4$

Using $x=-9$

$154-(-9)=163$
$50-2(-9)=68$
$(-9)+120=111$
$(-9)^2+7(-9)=18$

Now, you try using $x=4$ and see what happens.

magentarita · #4 $Old$ January 9th, 2009, 03:41 PM

I want to thank both replies.

masters · #5 $Old$ January 9th, 2009, 03:44 PM

Quote:

Originally Posted by magentarita $View Post$

I want to thank both replies.

Well.....was I right?!?

magentarita · #6 $Old$ January 9th, 2009, 08:16 PM

Quote:

Originally Posted by masters $View Post$

Well.....was I right?!?

Yes, you were right.

HallsofIvy · #7 $Old$ January 10th, 2009, 04:58 PM

Quote:

Originally Posted by vincisonfire $View Post$

Actually it is not a polygon since it has a curve in it.
Doing a sketch, you should easily find the angle and intersect of the lines.

I interpreted this to mean that these were the lengths of the sides, not equations of the lines. But I suspect that masters is right and they are really supposed to be measures of the angles so that they add to 360.

Quote:

$*50-2x = x +120$
$-70 = 3x$
$-\frac{70}{3} = x$
This is an example, it is NOT one of the vertexes.
For the parabola, I would use the derivative to find the slope at the intersect.
If needed ask for further explanations.

magentarita · #8 $Old$ January 12th, 2009, 11:25 PM

Yes, thank you all for your great help.