Question

The side of the cube is x units then the maximum length between any two vertices of the cube is

Answer 1

Formula we will be using:
(i) Pythagoras Theorem:
$c^2=a^2+b^2$, where a and b are the two legs of the right triangle and c is the hypotenuse.
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Please refer to the attached image, the longest diagonal is the red line.

Each side of the cube is x.
On of the diagonal to the bottom face is the green line.

The length of the green line is given by
 $c^2=a^2+b^2$
Plug in a=x and b=x:
$c^2=x^2+x^2$
$c^2=2x^2$
Square root both sides:
$c=x \sqrt{2}$

Therefore, length of the green line( bottom diagonal) is $x \sqrt{2}$ units.


Next, consider the right triangle constituted by the black line, green line and the red line.

The two legs of the right triangle -
The green line=$x \sqrt{2}$
The black line =x
And, the hypotenuse is the red line.

The length of the red line is given by,
$c^2=a^2+b^2$
Plug in a=x and b=$x \sqrt{2}$:
$c^2=x^2+(x \sqrt{2})^2$
$c^2=x^2+2x^2$
$c^2=3x^2$
Square root both sides:
$c=x \sqrt{3}$

Therefore, the maximum length between any two vertices of the cube is $x \sqrt{3}$ units

Answer : $x \sqrt{3}$ units