The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex of the chosen corner. What is the height of the remaining cube when the freshly-cut face is placed on a table?
+2862 
This is what the cutting is? Im just gonna solve it like that...
Now we turn the cube on its cut face (i suck at drawing so its hard to comprehend). Imagine that the cut face (triangle) is on the table.
Commence calculations
Sorry this may be EXTREMELY difficult to comprehend because I suck at drawing lol
Hypotenuse = \(\sqrt{2}\)
Long Leg = \(\frac{\sqrt{6}}{3}\)
By pythageoran theorem:
Answer = \(\sqrt{(\sqrt{2})^2-(\frac{\sqrt{6}}{3})^2}\)
We get a radical of
\(\boxed{\frac{2\sqrt{3}}{3}}\)
+2862 [previous deleted] ok imma gonna do dis tommorow because it is 10:30 for me now
+2862
This is what the cutting is? Im just gonna solve it like that...
Now we turn the cube on its cut face (i suck at drawing so its hard to comprehend). Imagine that the cut face (triangle) is on the table.
Commence calculations
Sorry this may be EXTREMELY difficult to comprehend because I suck at drawing lol
Hypotenuse = \(\sqrt{2}\)
Long Leg = \(\frac{\sqrt{6}}{3}\)
By pythageoran theorem:
Answer = \(\sqrt{(\sqrt{2})^2-(\frac{\sqrt{6}}{3})^2}\)
We get a radical of
\(\boxed{\frac{2\sqrt{3}}{3}}\)
