Question 1151099
 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}}}. and it is 
{{{TZ }}}= solid diagonal

Again, by the Pythagorean theorem we know that

{{{(TZ)^2 = (WZ)^2 + (WT)^2}}}
{{{(4sqrt(3 )cm)^2 = (WZ)^2 + (WT)^2}}}........since cube, all sides are equal, so {{{WV=WU=WZ}}} and {{{(WT)^2=(WV)^2+(WU)^2=(WZ)^2+(WZ)^2=2(WZ)^2}}}

{{{16*3cm^2 = (WZ)^2 + 2(WZ)^2}}}

{{{16*3cm^2 = 3(WZ)^2}}}
{{{(16*3cm^2)/3 = (WZ)^2}}} 

{{{(WZ)^2=16cm^2}}}

{{{WZ=4cm}}}........since cube, {{{ZY=4cm}}}

point {{{X}}} is the intersection of the two diagonals,{{{ XZ=XY}}}

triangle is isosceles, and altitude from {{{X}}} to {{{ZY}}} will bisect{{{ ZY}}}, let say at point {{{M}}} and {{{XM}}} divides triangle {{{XYZ}}} into two right triangles

 altitude from {{{X }}}to {{{WV}}} will bisect {{{WV}}}, let say at point {{{N}}}, and the length of {{{XN =2cm}}}

now, connect points {{{N}}} and {{{M}}} and you got right triangle {{{XNM}}} whose side {{{XM}}} is altitude from {{{X}}} to {{{ZY}}}

sides of this right triangle are:

 {{{XN =2cm}}}
{{{NM=WZ=4cm}}}

find altitude {{{XM}}}:

{{{(XM)^2=(2cm)^2+(4cm)^2}}}
{{{(XM)^2=4cm^2+16cm^2}}}
{{{(XM)^2=20cm^2}}}
{{{XM=sqrt(20cm^2)}}}
{{{XM=sqrt(4*5)cm}}}
{{{XM=2sqrt(5)cm}}}

then, the area   of triangle {{{XYZ}}} will be:

{{{A=(1/2)ZY*XM}}}

{{{A=(1/2)4cm*2sqrt(5)cm}}}

{{{A=(1/cross(2))4cm*cross(2)sqrt(5)cm}}}

{{{A=4sqrt(5)cm^2}}}