New Math Problem: Paper Folding

[gentry_joey]

Take an ordinary piece of typing paper; its dimensions will be 8 1/2 inches by 11 inches. Label the four corners A, B, C, and D clockwise from the top left (so that the short sides of the paper are AB and CD ). Label the midpoint of AB as E and the midpoint of BC as F. Now draw straight lines from A to F and from D to E. Label the point at which these lines cross as G. Now, find the exact area of the four-sided region GFCD. (No calculator approximations will count!) Be sure to explain how you got your answer.

[John Prophet]

is it yellow paper or white?

JP

[MrPurple]

Basicaly you find the angles of triangle ABF, or ADE, and using trig rules you find the angles. Once you find them you will get the angles of the small triangle AGE. Using trig you find the sides and the area since it is a 90 degree triangle.
Finaly you find the total area of the rectangle, subtract the areas of triangles ABF and ADE, than add the area of triangle AGE (you subtracted it twice when you took out ABF and ADE), and you have your area.

I would post the answer here, but since I have not used trig in so long, I am having problems with the whole degree, radian settings, and rembering the rules.
But the theory is sound and should work.

Will post what I found out latter.

[The Real Bingo]

Shouldnt this topic be in the community?

[gentry_joey]

i just put it in general sorry.

[gentry_joey]

I dont think those angles are right angles.

[MrPurple]

They are, if you use cos, sin and tan rules on the two larger ones which defenatly have one 90 deg angle you will get the other two. With the other two you can get the angles of the smaller triangle in the top right corner (forgot the designation), and you will find that one angle is a 90 deg.

[SPEEDO]

Yeah but you need to know what the hypotenuse is to find the angle that's not 90 deg. or the angle itself.

[MrPurple]

you can get all the angles and dimentions of the two larger triangles (they are identical BTW).
You have the base and height, and it is a 90 deg traiangle, so you use a^2+b^2=c^2 where a and b are height and base, while c is the hypotenuse.
Once you get the angles of the smaller triangle you can use trig to find the missing lenghts, since you have one of them, and it is the hypotenuse of that triangle. The angles will show this to be true

PS I would like to thank you all for allowing me to post my 400 post

[SPEEDO]

OK I'll bite...................

So what's the answer that he asked for at the beginning of this thread?

I understand what your saying but just give me an angle besides the 90 deg one..............