Super Pythagorean???

[KTsincere]

Can someone explain the Super Pythagorean (when to use it, why, how)????
I'm a bit confused....
Thanks a lot

[Sudhan]

The super Pythagorean Theorem refers to the Law of Cosines which says for any triangle with sides a,b,c and opposite angles A,B, and C then

c^2 = a^2 + b^2 -2ab(cosC)

If C=90 degrees then cos90 = 0 which means the equation becomes Pythagorean Theorem.

You can think of the law of cosines as the Pythagorean Theorem for any triangle with a correction factor for non-right triangles of -2ab(cosC).

It depends on size and position of sphere in cube. assuming the sphere fits the cube, or even if the sphere is located at cube centre, let cube side be x. then the distance is x/2*Sqrt3.

Source: Yahoo


Hope it Helps.

[RonPurewal]

the rule in sudhan's post is for real, but it's not anything you'll need for the gmat. the gmat doesn't require any trigonometry; the closest you'll get is 30-60 and 45-45 right triangles, but those can be handled via memorized templates (i.e., you don't have to use trig functions to solve them if you can remember the 1-1-root(2) and 1-root(3)-2 templates).

--

presumably, 'super pythagorean' refers to the following rule (which i myself refer to as 'p.t. deluxe' in class - sudhan, you should remember this if you're the same sudhan who took my virtual class recently):

if D is the MAIN DIAGONAL OF A BOX whose length is L, width is W, and height is H, then:
D^2 = L^2 + W^2 + H^2
or,
D = root(L^2 + W^2 + H^2)

notice that the first version is exactly the same as the pythagorean theorem with one additional term tacked onto it; hence the name 'super pythagorean' or 'p.t. deluxe'.

if you ever need to find a diagonal distance in three dimensions, this is the formula to use.

[Sudhan]

Thanks Ron. Now I remember you taught this formula in the class.

[KTsincere]

Thanks for the clarification... So just to make sure I understand, the main diagonal of a box is the longest diagonal correct?

[Sudhan]

yes.

[RonPurewal]

Thanks for the clarification... So just to make sure I understand, the main diagonal of a box is the longest diagonal correct?

yeah.

you might know this already, but, full disclosure: there are actually 4 main diagonals, all of which have the same length. it's difficult to describe this without a diagram, but imagine that the sides of the box face north, east, south, and west. then the main diagonals run from:
top NW to bottom SE
top NE to bottom SW
top SW to bottom NE
top SE to bottom NW
they're all the same length, and they're all longer than anything else you can draw in the box.

[gkhan]

the rule in sudhan's post is for real, but it's not anything you'll need for the gmat. the gmat doesn't require any trigonometry; the closest you'll get is 30-60 and 45-45 right triangles, but those can be handled via memorized templates (i.e., you don't have to use trig functions to solve them if you can remember the 1-1-root(2) and 1-root(3)-2 templates).

--

presumably, 'super pythagorean' refers to the following rule (which i myself refer to as 'p.t. deluxe' in class - sudhan, you should remember this if you're the same sudhan who took my virtual class recently):

if D is the MAIN DIAGONAL OF A BOX whose length is L, width is W, and height is H, then:
D^2 = L^2 + W^2 + H^2
or,
D = root(L^2 + W^2 + H^2)

notice that the first version is exactly the same as the pythagorean theorem with one additional term tacked onto it; hence the name 'super pythagorean' or 'p.t. deluxe'.

if you ever need to find a diagonal distance in three dimensions, this is the formula to use.

Ron,
Just saw forum rules, moved q to a new thread, sorry about that.

[RonPurewal]

Ron,
Just saw forum rules, moved q to a new thread, sorry about that.

thanks