# Thread: Geometry: Porportions and Area of regular Polygons -Help!

1. ## Geometry: Porportions and Area of regular Polygons -Help!

Hey im new here and I need some math help FAST!

Is there anyone who can solve these and show the work?
Im kind of confused...

Sorry if its sloppy,

<Drumm3r>

2. just to clarify:

in the first question what are you trying to find? the area of the large figure minues the area of the small figure?

3. if the polygons are regular then the distance from the vertices to the center are equal

further you can solve for the angle between the line segment joining the vertices and the center

edit: (this is the angle at the center that is contained by the two line segments)
for the pentagon the angle would be 360/5 in degrees
and for the hexagon the angle would be 360/6 degrees

from here you can use trigonometric ratios, pythagorean theorem and area formula for triangles to solve for the area of each

in the second one you have the square with length of center to vertices is 8
this makes 4 isosceles triangles with angles 45, 45, 90
you can find the angles that are 90 by doing a similar trick as above: 360/4

so you have a right triangle with length 8 and 8 and you can solve for the hypotenuse to get the length of one side of the square: thus you can get the area

the area of the circle is (pi)(r squared)

you can find the probability of hitting the square by doing the ratio of the square area to the triangle area

area square/area triangle x 100 = %
then you can find the probability of hitting the shaded area by taking 100% - the calculated probablity of hitting the square

i hope this helps in some way or another

4. For the first one, just both area's of both figures...

like the area of the small one is : x
and the area of the big polygon is : x

I could finish the problem from there...

+rep to you, since you helped...

<Drumm3r>

5. Probability of hitting the shaded region is proportional to the area. So take the ratio of the areas.

For starters, the side of the square is $8\sqrt2$ by Pythogorus theorem.