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# The fastest way?.

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12 Dec 2006, 19:43
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The fastest way?..
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Senior Manager
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12 Dec 2006, 21:13
the fastest way? not sure....
i used pythagoras theorem twice...

however, to save calculation time, i divided everything by 4.
(so the sides are 3,4,5). now, some of you know by heart that 3^2+4^2=5^2 (the most known example)...
so 3 and 4 gives 5.

5 and 5 gives 5*sqrt(2). or you may now by heart (another common example) that 1 and 1 give sqrt(2). so divide by 5.

so after first dividing by 4, then dividing by 5 we get sqrt(2) then the answer is 20*1.41 ~28

you could of course use just one step, knowing that in three dimensional space l^2 = a^2+b^2+c^2

but why bother remembering so much stuff...

however - the key to fast calculation, in my view, is breaking stuff into known and recognizable patterns, do the fast calculations on them, then transform it back.

!!!!better 10 fast and easy steps than 1 complicated step.!!!!

this at least was (and still is) my way of approaching stuff.

amit.
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12 Dec 2006, 21:27
Yes 28

D = sqrt(16^2 + 12^2 + 20^2) = sqrt (800) = sqrt(400 * 2) = 20 sqrt2
= 20 * 1.41.. > 28
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12 Dec 2006, 21:37
I used the formula :

OA is 28

Thanks!
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12 Dec 2006, 21:41
hobbit wrote:
the fastest way? not sure....
i used pythagoras theorem twice...

however, to save calculation time, i divided everything by 4.
(so the sides are 3,4,5). now, some of you know by heart that 3^2+4^2=5^2 (the most known example)...
so 3 and 4 gives 5.

5 and 5 gives 5*sqrt(2). or you may now by heart (another common example) that 1 and 1 give sqrt(2). so divide by 5.

so after first dividing by 4, then dividing by 5 we get sqrt(2) then the answer is 20*1.41 ~28

you could of course use just one step, knowing that in three dimensional space l^2 = a^2+b^2+c^2

but why bother remembering so much stuff...

however - the key to fast calculation, in my view, is breaking stuff into known and recognizable patterns, do the fast calculations on them, then transform it back.

!!!!better 10 fast and easy steps than 1 complicated step.!!!!

this at least was (and still is) my way of approaching stuff.

amit.

In doing such questions, I have noticed that we would often need to figure out the length of the longest diagonal of the cube.

I simply do this.

length=sqrt[(longestside^2)+(sqrt(sum of squares of the other 2 sides))^2]

It's still the same formulae, but just that I dont bother about visualizing or drawing the cube.
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13 Dec 2006, 11:16
C - 28

I used ratios 3:4:5 then x:x:x(sqrt(2))
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15 Dec 2006, 20:45
C

12, 16 >> 20
20,20 >>sqrt2 * 20 = 1.414 *20 =apprx 28
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15 Dec 2006, 22:10
basically the q is asking the length of the diagnonal in a 3-d frame.for which u have to take the diagonal of one face(Sqrt(a^2+b^2)=d
and then the diagonal of d and the 3rd side c.

in nutshell,find sqrt(a^2+b^2+c^2)

which is 20sqrt2 = 28
Re: PS - rod   [#permalink] 15 Dec 2006, 22:10
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