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
Find the area in square centimetres of triangle XYZ.
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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. In this problem, you are given that the longest 3D-diagonal of the cube is 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 = = = cm. Now the area of the triangle XYZ is half the product of its base a = ZY = 4 cm by the altitude h = = = 8.944 cm^2 (approximately). ANSWER
Solved.