Question 1151099: In the diagram below, 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 4 sqrt3 cm. Find the area in cm2, of triangle XYZ.
Sry bout the other one.
Diagram: https://imgur.com/a/ciK5pIK
Answer by MathLover1(20849)   (Show Source):
You can put this solution on YOUR website! point  is the intersection of the two diagonals  and  of the cubical box illustrated.
The shortest distance from point  to point  is  . and it is
 = solid diagonal
Again, by the Pythagorean theorem we know that
 ........since cube, all sides are equal, so  and 
 ........since cube, 
point is the intersection of the two diagonals,
triangle is isosceles, and altitude from to  will bisect , let say at point  and  divides triangle  into two right triangles
altitude from  to  will bisect , let say at point  , and the length of 
now, connect points and and you got right triangle  whose side is altitude from to
sides of this right triangle are:
find altitude :
then, the area of triangle will be:
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