SFO Archive Forums - View Single Post - A little algebra ii

Saijin_Ryu · 11-07-2004, 11:20 AM

[quote

ost_uid0="TarkanX"]Like a few have explained, since the radius of the circle is one, the diameter of the circle is two, and if the diameter of the circle is two, then the hypotenuse of the sliced square (cut diagnolly) is two. So it would be:

a^2 + b^2 = c^2

= a^2 + b^2 = 2^2

a = b, so...

(2a^2 = 4) / (2)

= a^2 = 2

Square root both and you get:

a = square root of 2, let's see if this is correct.

v - represents square root of.

v2^2 + v2^2 = 2^2

2 + 2 = 4.

We are right, so the length and width would be v2, which is approximately 1.4142.[/quote]
The bits in italic is what i understood up too.

11-07-2004, 11:20 AM	#16
Saijin_Ryu Senior Member Senior Member Join Date: Mar 2003 Location: Somewhere in England Posts: 4,082 Post Thanks / Like Thanks (Given): Thanks (Received): Likes (Given): Likes (Received): Dislikes (Given): Dislikes (Received): Points: 0 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s)	[quoteost_uid0="TarkanX"]Like a few have explained, since the radius of the circle is one, the diameter of the circle is two, and if the diameter of the circle is two, then the hypotenuse of the sliced square (cut diagnolly) is two. So it would be: a^2 + b^2 = c^2 = a^2 + b^2 = 2^2 a = b, so... (2a^2 = 4) / (2) = a^2 = 2 Square root both and you get: a = square root of 2, let's see if this is correct. v - represents square root of. v2^2 + v2^2 = 2^2 2 + 2 = 4. We are right, so the length and width would be v2, which is approximately 1.4142.[/quote] The bits in italic is what i understood up too.