Need help finding diagonal of rectangular box in terms of k

lp_144

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Jul 14, 2008
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Can someone explain to me how to solve this problem?
 
Re: Need help solving this problem.

for a rectangular prism with dimensions l, w, and h, the diagonal length is
\(\displaystyle \sqrt{l^2 + w^2 + h^2}\)
 
Re: Need help solving this problem.

skeeter said:
for a rectangular prism with dimensions l, w, and h, the diagonal length is
\(\displaystyle \sqrt{l^2 + w^2 + h^2}\)

Ohh, that helps a lot.
So the answer would be \(\displaystyle \sqrt{2 + k^2}\).
Thanks! :mrgreen:
 
I have a tough time NOT responding to this question. It seems to me that the questioner is content to use the given formula without understanding why it works. I would encourage him/her to draw a diagonal on the side face of the cube and calculate it's measure --- possibly \(\displaystyle \sqrt{k^2+1^2}\). Then he/she could work with the inner right triangle where AB is the hypotenuse. The result would be \(\displaystyle (AB)^2=(\sqrt{k^2+1^2} )^2 + 1^2\). From there he/she could arrive at the final result.
 
Loren said:
I have a tough time NOT responding to this question. It seems to me that the questioner is content to use the given formula without understanding why it works. I would encourage him/her to draw a diagonal on the side face of the cube and calculate it's measure --- possibly \(\displaystyle \sqrt{k^2+1^2}\). Then he/she could work with the inner right triangle where AB is the hypotenuse. The result would be \(\displaystyle (AB)^2=(\sqrt{k^2+1^2} )^2 + 1^2\). From there he/she could arrive at the final result.

Loren, I don't understand what you are telling me to do. :( I understand the formula skeeter provided me with because it's simple. Can you please elaborate with what you are trying to suggest? I want to understand your method as well.
 
Loren is trying to tell you that it is important that you know HOW that formula was arrived at.

If you label the corner under B as C, then join AC, you get right triangle ABC; see that?
So AB^2 = AC^2 + BC^2

Also, AC is the hypotenuse of a right triangle with legs k and 1; see that?

So AC^2 = k^2 + 1^2
And since BC = 1 .... can you wrap it up?
 
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