Surface Area of a Truncated cube

• May 3rd 2009, 03:10 PM
algebraisabeast
Surface Area of a Truncated cube
So I have a cube that is 1X1 so the surface area of the cube will be 6 I want to know how much the surface area will be changed if I cut off the edge, the truncated cube is 5.37 sq in What I cut off will be a triangular pyramid If I find the area of each side I figure it to be 1.5 units taken off not .63 Help on how to get the .63 of how to get the answer would be great thanks
• May 3rd 2009, 09:57 PM
Truncated cube
Hello algebraisabeast
Quote:

Originally Posted by algebraisabeast
So I have a cube that is 1X1 so the surface area of the cube will be 6 I want to know how much the surface area will be changed if I cut off the edge, the truncated cube is 5.37 sq in What I cut off will be a triangular pyramid If I find the area of each side I figure it to be 1.5 units taken off not .63 Help on how to get the .63 of how to get the answer would be great thanks

As the pyramid is cut off it reveals a new triangular surface whose area must now be included in the total surface area of the new solid.

If the cut is made through three of the cube's vertices (which I assume to be the case), then, as you say, the area removed is $3 \times 0.5 = 1.5$. But the new area revealed is an equilateral triangle whose sides are $\sqrt2$ units long, and whose area (using area of triangle = $\tfrac12 bc\sin A$) is

$\tfrac12\sqrt2\sqrt2\sin60^o = 0.866$

So the new area is $6 -1.5+0.866 = 5.37$ (to 2 d.p.)