Question 1028361
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A cube has a diagonal.  I don't know what a "diagnol" is.


If the edge of a cube is represented by *[tex \Large x], then the diagonal of one face is *[tex \Large x\sqrt{2}].  This is from the Pythagorean Theorem, *[tex \Large d\ =\ \sqrt{x^2\ +\ x^2}].


Then one edge of the cube, measuring *[tex \Large x] and the diagonal of a face, measuring *[tex \Large x\sqrt{2}], form the legs of another right triangle where the diagonal of the cube is the hypotenuse.  So using the Pythagorean Theorem again:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ D\ =\ \sqrt{(x\sqrt{2})^2\ +\ x^2}\ =\ x\sqrt{3}]


So, if your diagonal measures *[tex \Large 12], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\sqrt{3}\ =\ 12]


Solving and rationalizing the denominator


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 4\sqrt{3}]


The volume of the cube is the measure of the edge cubed, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(4\sqrt{3}\right)^3]


You can do the rest of the arithmetic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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