Question 1205618
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The space diagonal connects two opposite corners of a cube. 
One example of a space diagonal is us connecting points T and Z. Think of a rope going through a room. 


In contrast, segments TW and VU are face diagonals since they are contained entirely on one face or plane.
These two ropes are glued entirely to one wall.
The face diagonals intersect at point X. 
It turns out that this is the midpoint of segments TW and VU. 
I'll leave the proof for the reader to determine.


All sides of a cube are the same length.
s = side length of the cube


Draw a segment to connect points V and Z.
Focus on right triangle VWZ.
leg1 = VW = s
leg2 = WZ = s
hypotenuse = VZ = unknown
Use the Pythagorean theorem to determine that hypotenuse VZ = s*sqrt(2)
I'll leave the steps and scratch work for the student to do.


Move your focus to right triangle TVZ.
leg1 = TV = s
leg2 = VZ = s*sqrt(2)
Use the Pythagorean theorem to determine that hypotenuse TZ = s*sqrt( 3 ).
I'll leave the steps and scratch work for the student to do.
Because we're told that TZ = 4*sqrt(3), this must mean that s = 4.
The cube has a side length of 4 units.


Define these two new points
P = midpoint of YZ
Q = midpoint of VW


Right triangle PQX has these side lengths
PQ = horizontal leg = 4
QX = vertical leg = 2 (since X is at the midpoint)
PX = unknown hypotenuse


Use the Pythagorean theorem one last time
(PQ)^2 + (QX)^2 = (PX)^2
(4)^2 + (2)^2 = (PX)^2
(PX)^2 = 20
I skipped a few steps and will let the student fill in the details.
PX = 2*sqrt(5)


Triangle XYZ has base YZ = 4 and height PX = 2*sqrt(5)


area = 0.5*base*height
area = 0.5*YZ*PX
area = 0.5*4*2*sqrt(5)
<font color=red>area = 4*sqrt(5) square cm</font>
4*sqrt(5) = 8.944271909999 approximately 
Round this however needed.
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