Question 1205620
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<img src="https://i.ibb.co/P1MwGxB/Untitled-drawing.png" height="200px">

Point X is the intersection of the two diagonals TW and UV of the cubical box illustrated. 
The shortest distance from point T to point Z is {{{4sqrt(3)}}} cm. 
Find the area in square centimetres of triangle XYZ.
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<pre>
It is well known fact that if "a" is the length of the edge of a cube,
then the longest 3D-diagonal of the cube, connecting the opposite corners (vertices)
is  {{{a*sqrt(3)}}}.


In this problem, you are given that the longest 3D-diagonal of the cube is {{{4*sqrt(3)}}} cm.
It means that the edge of the cube is 4 cm long.


OK.  To find the area of triangle XYZ, we need to know its height (its altitude).
This altitude is the hypotenuse of the right angled triangle with the legs a/2 = 2 cm
and a = 4 cm, so the altitude of the triangle XYZ is

    h = {{{sqrt(2^2 + 4^2)}}} = {{{sqrt(20)}}} = {{{2*sqrt(5)}}} cm.


Now the area of the triangle XYZ is half the product of its base a = ZY = 4 cm by the altitude h

    {{{area[XYZ]}}} = {{{(1/2)*4*2*sqrt(5)}}} = {{{4*sqrt(5)}}} = 8.944 cm^2  (approximately).    <U>ANSWER</U>
</pre>

Solved.