# Thread: Polygon Sides

1. ## Polygon Sides

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

2. 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.

3. Originally Posted by magentarita
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.

4. ## Thanks

I want to thank both replies.

5. Originally Posted by magentarita
I want to thank both replies.
Well.....was I right?!?

6. ## yes...

Originally Posted by masters
Well.....was I right?!?
Yes, you were right.

7. Originally Posted by vincisonfire
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.

$*50-2x = x +120$
$-70 = 3x$
$-\frac{70}{3} = x$
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.

8. ## yes......

Yes, thank you all for your great help.