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# The corner of a unit cube is chopped off such that the cut runs through the three vertices adjacent to the vertex

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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?

Feb 12, 2020

#3
+2856
+3

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 = $$\sqrt{(\sqrt{2})^2-(\frac{\sqrt{6}}{3})^2}$$

$$\boxed{\frac{2\sqrt{3}}{3}}$$

Feb 13, 2020
edited by CalculatorUser  Feb 13, 2020

#1
+2856
0

[previous deleted] ok imma gonna do dis tommorow because it is 10:30 for  me now

Feb 12, 2020
edited by CalculatorUser  Feb 12, 2020
edited by CalculatorUser  Feb 12, 2020
#2
+110206
0

What does that even mean?

Feb 12, 2020
#3
+2856
+3

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 = $$\sqrt{(\sqrt{2})^2-(\frac{\sqrt{6}}{3})^2}$$

$$\boxed{\frac{2\sqrt{3}}{3}}$$

CalculatorUser Feb 13, 2020
edited by CalculatorUser  Feb 13, 2020
#4
+2856
+2

bruh moderation

CalculatorUser  Feb 13, 2020
#5
+110206
+3

Thanks CalculatorUser!

Very nicely executed!

Melody  Feb 13, 2020