Difference between revisions of "Rectangular prism"

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A '''rectangular prism''' (also '''cuboid''', '''rectangular box''', '''right rectangular prism''', '''rectangular paralleliped''') is a [[3D|three dimensional]] figure with 6 [[face]]s that are all [[rectangle]]s.  
 
A '''rectangular prism''' (also '''cuboid''', '''rectangular box''', '''right rectangular prism''', '''rectangular paralleliped''') is a [[3D|three dimensional]] figure with 6 [[face]]s that are all [[rectangle]]s.  
  
Opposite faces of a rectangular prism are [[congruent]] and [[parallel]].
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Opposite faces of a rectangular [[prism]] are [[congruent (geometry) | congruent]] and [[parallel]].
  
The [[volume]] can be determined by multiplying the length, width, and height (<math>V = lwh</math>).  
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The [[volume]] can be determined by multiplying the length, width, and height, <math>V = lwh</math>.  
  
The length of a diagonal can be determined by using the formula <math>\sqrt{l^2 + w^2 + h^2}</math>.
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The length of the interior [[diagonal]]s can be determined by using the formula <math>d = \sqrt{l^2 + w^2 + h^2}</math>.
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Proof: To get a base diagonal, we use the [[pythagorean theorem]]: <math> \sqrt{l^2+w^2}</math>. We call that  v. Then we use the pythagorean theorem again to get
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* <math>diagonal=\sqrt{v^2+h^2}=\sqrt{l^2+w^2+h^2}</math>
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* The [[surface area]] of the prism is <math>2lw+2wh+2lh</math>
  
 
==See also==
 
==See also==
*[[Cube]]
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*[[Cube (geometry) | Cube]]
  
 
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Latest revision as of 09:04, 12 September 2007

A rectangular prism (also cuboid, rectangular box, right rectangular prism, rectangular paralleliped) is a three dimensional figure with 6 faces that are all rectangles.

Opposite faces of a rectangular prism are congruent and parallel.

The volume can be determined by multiplying the length, width, and height, $V = lwh$.

The length of the interior diagonals can be determined by using the formula $d = \sqrt{l^2 + w^2 + h^2}$.

Proof: To get a base diagonal, we use the pythagorean theorem: $\sqrt{l^2+w^2}$. We call that v. Then we use the pythagorean theorem again to get

  • $diagonal=\sqrt{v^2+h^2}=\sqrt{l^2+w^2+h^2}$


See also

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