Anyone know Trigonometry want to help me on a simple question?

[KAMAZON]

I was having a discussion on how to double a square which has an area of 1, as discussed in Platos mino dialogue. The method that is proven to work is to bisect the square into 2 triangles, and use the hypotenuse as the side of the new square which has exactly an area of 2. I wanted to see if this solution is also correct. Can anyone tell me the length of the Hypotenuse in this picture?

http://www.chamberband.com/ali/images/square.jpg

The length/width/area of the original black square is 1. Thus each side is 1. Thus the bisected triangle at the north of the square have a base length of .5 (lets say .5 meters, or whatever metric value you want to give it). So...This is a 45 degree triangle bisected into 2 90 degree triangles. Can anyone give me the length of the hypotenuse of the triangles? I'm trying to see if this method also doubles the square or not. Thank you in advance.



EDIT: Fixed spelling for all the english majors that don't know math.

 

[mordantmonkey]

lol, hypatnuse. you need more than trig help little hombre.

 

[olds]

Take English first

 

[KAMAZON]

Last time I took trig was over 5 years ago, and never used it untill i started reading books for 2100 years ago. Go figure.

 

[KAMAZON]

Originally posted by: oldsmoboat
Take English first

Oh noes, I spelled a latin word wrong!

 

[MazerRackham]

Originally posted by: oldsmoboat
Take English first

ROFLMAO! You just made me spit out my drink... thank a lot!

 

[Juice Box]

double post

 

[sao123]

the hypotenuse in any rt triangle is always

C = sqrt(A^2 + B^2)
= sqrt( .5 * .5 + .5 * .5)
= sqrt( .25 + .25)
= sqrt( .5) ~~ .70710678118654

 

[KAMAZON]

Ok that's the figure I also got, which means that this method although looks like it works, does NOT double the square.

 

[chuckywang]

it's sqrt(2)/2.

 

[mordantmonkey]

you are essentially doing the same thing as the first solution. the hypatnuse(sic) of those triangles is one half the squareroot of 2. the original solution creates four triangles with the hypotenuse of square root of 2 instead of sixteen with hyp of 1/2 that.

 

[VTHodge]

Yes, you are correct. Imagine taking two squares, cutting them along the 45 deg angle and aranging the four new triangles into one big squaure, with all the old 90 deg angles at the new center point. This square has twice the area of two original squares and a side length of sqrt(2), which was the hypotaneuse of the original square.

The length of the hyp. of a small blue triangle in the picture is 0.5*sqrt(2). Imagin mirroring it along the top-most horizontal line. It will be half the lenght of the black square's hyp.

Hope this helps.

 

[OulOat]

It does double the area.

Proof:
a = Orginal Square Length
a*a = a^2 = Original Square Area

c = Hypotenuse of Orginal Square and the Length of the Side in the Larger Square
c^2 = a^2 + a^2 (Pythagorean Theorem a=b since this is a square) = 2a^2

c*c = c^2 = Area of Larger Square

c^2 / a^2 = How many times larger the Larger Square is compared to the Orginal Square
c^2 / a^2 = (2a^2) / (a^2) = 2

EOD

 

[KAMAZON]

Originally posted by: OulOat
It does double the area.

Proof:
a = Orginal Square Length
a*a = a^2 = Original Square Area

c = Hypotenuse of Orginal Square and the Length of the Side in the Larger Square
c^2 = a^2 + a^2 (Pythagorean Theorem a=b since this is a square) = 2a^2

c*c = c^2 = Area of Larger Square

c^2 / a^2 = How many times larger the Larger Square is compared to the Orginal Square
c^2 / a^2 = (2a^2) / (a^2) = 2

EOD

God dang I wish I remembered Trig grrr. I'm confused now. SAO 123 states:
C = sqrt(A^2 + B^2)
= sqrt( .5 * .5 + .5 * .5)
= sqrt( .25 + .25)
= sqrt( .5) ~~ .70710678118654


If the length of the hyp is .70, then there is no way it would create a square with an area of 2. So are you stating that Saos equation is wrong for this example?

 

[OulOat]

Originally posted by: KAMAZON

Originally posted by: OulOat
It does double the area.

Proof:
a = Orginal Square Length
a*a = a^2 = Original Square Area

c = Hypotenuse of Orginal Square and the Length of the Side in the Larger Square
c^2 = a^2 + a^2 (Pythagorean Theorem a=b since this is a square) = 2a^2

c*c = c^2 = Area of Larger Square

c^2 / a^2 = How many times larger the Larger Square is compared to the Orginal Square
c^2 / a^2 = (2a^2) / (a^2) = 2

EOD

God dang I wish I remembered Trig grrr. I'm confused now. SAO 123 states:
C = sqrt(A^2 + B^2)
= sqrt( .5 * .5 + .5 * .5)
= sqrt( .25 + .25)
= sqrt( .5) ~~ .70710678118654


If the length of the hyp is .70, then there is no way it would create a square with an area of 2. So are you stating that Saos equation is wrong for this example?

He used the specific case (one number) of the side in the orginal square length = .5, hence A = B = .5.
Orginal Area = A*B = .5 * .5 = .25
Larger Area = C * C = sqrt( .5) * sqrt( .5) = sqrt( .5) ^ 2 = .5

Multiple in Areas = .5 / .25 = 2

Mine is a general case (all numbers), since I just use variables instead of numbers.

If you want do a quick check, a specific case is a quick and easy way. But if you want to actually prove it, you have to use the general case.

 

[sao123]

Originally posted by: KAMAZON

Originally posted by: OulOat
It does double the area.

Proof:
a = Orginal Square Length
a*a = a^2 = Original Square Area

c = Hypotenuse of Orginal Square and the Length of the Side in the Larger Square
c^2 = a^2 + a^2 (Pythagorean Theorem a=b since this is a square) = 2a^2

c*c = c^2 = Area of Larger Square

c^2 / a^2 = How many times larger the Larger Square is compared to the Orginal Square
c^2 / a^2 = (2a^2) / (a^2) = 2

EOD

God dang I wish I remembered Trig grrr. I'm confused now. SAO 123 states:
C = sqrt(A^2 + B^2)
= sqrt( .5 * .5 + .5 * .5)
= sqrt( .25 + .25)
= sqrt( .5) ~~ .70710678118654


If the length of the hyp is .70, then there is no way it would create a square with an area of 2. So are you stating that Saos equation is wrong for this example?

ERROR: the error is in the picture... you drew on the wrong line... thats why its confusing the sides are not .5 & .5, they are 1 & 1.

C = sqrt(A^2 + B^2)
= sqrt( 1 * 1 + 1 * 1)
= sqrt( 1 + 1)
= sqrt( 2) ~~ 1.41421....


1.41421...... = sqrt(2)

Area of new square.
sqrt(2) * sqrt(2) =

2



Correctly Drawn Picture
This one has the triangle highlighted

 

[KAMAZON]

Well the base of the origional black square was 1. Since I spilt it into 2 right angle triangles, the base of the new triangles each is .5. So is it still an area of 2?

 

[Leper Messiah]

nm, im stupid.

 

[akubi]

um of course the area is still 2.
look at the picture you drew. what is the length of a side of the purple square? it's the same as the length of the diagnal of the black square.

so both "methods" ( :roll: ) yield the "same square" ( :roll: ), just "shifted" ( :roll: ) by half its width...

(edit: fixed eye roll emoticons :laugh

 

[KAMAZON]

Everyone thank you for responding.