Question 1189507: In the diagram below, Point X is the intersection of the two diagonals TW and UV of the cubical box illustrated. Given that UX is 2 √2 cm, what is the square of the area of triangle XYZ in cm^4
Diagram: https://imgur.com/a/miA6pQf
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Given info:
X is the intersection of lines TW and UV
UX = 2*sqrt(2)
The 3D figure is a cube, which has 6 square faces.
That means the 2D quadrilateral VWUT is a square.
Draw out VWUT on a separate sheet of paper or off to the side.
Include the diagonals TW and UV, along with the point X in the middle.
Recall that a square has its diagonals bisect each other.
This property applies to rectangles as well (any square is a rectangle but not vice versa).
So this means UX = XV = 2*sqrt(2)
We can then say:
UV = UX+XV
UV = 2*sqrt(2) + 2*sqrt(2)
UV = 4*sqrt(2)
The other diagonal TW is the same length as UV.
Now focus on right triangle VWU.
We found the hypotenuse to be UV = 4*sqrt(2)
Let's say the side lengths of the cube are x units
So VW = x and UW = x also.
Apply the pythagorean theorem to find x.
a^2 + b^2 = c^2
(VW)^2 + (UW)^2 = (UV)^2
x^2+x^2 = (4*sqrt(2))^2
2x^2 = 32
x^2 = 32/2
x^2 = 16
x = 4
The sides of the cube are 4 cm.
This means YZ = 4 which we'll use later. This is the base of triangle XYZ.
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Let's add two points to this diagram.
Plot point A directly below point X such that point A is on segment VW.
Plot point B on segment YZ that is directly across from point A.
We'll make segment AB to be parallel to VY and WZ.
By construction, points A and B are midpoints of VW and YZ respectively.
Turn your attention to triangle BAX.
This is a right triangle with legs of XA = 2 (half the cube's side length) and AB = 4.
The hypotenuse BX is the height of triangle XYZ.
We'll use the pythagorean theorem again.
a^2 + b^2 = c^2
(XA)^2 + (AB)^2 = (BX)^2
BX = sqrt( (XA)^2 + (AB)^2 )
BX = sqrt( (2)^2 + (4)^2 )
BX = sqrt( 20 )
BX = sqrt(4*5)
BX = sqrt(4)*sqrt(5)
BX = 2*sqrt(5)
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We found that:
YZ = 4
BX = 2*sqrt(5)
which represent the base and height of the triangle XYZ.
We conclude things by using the formula below
area = (1/2)*base*height
area = (1/2)*YZ*BX
area = (1/2)*4*2*sqrt(5)
area = 4*sqrt(5)
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Final Answer:
4*sqrt(5) square cm
This value is exact.
If you were to use your calculator to compute the approximate area, then you'll get roughly 4*sqrt(5) = 8.94427
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