SOLUTION: Here is a solution by MathLover1:
http://www.algebra.com/tutors/students/your-answer.mpl?question=1205618
I would never in a million years trust her solution, given her reputat
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-> SOLUTION: Here is a solution by MathLover1:
http://www.algebra.com/tutors/students/your-answer.mpl?question=1205618
I would never in a million years trust her solution, given her reputat
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Question 1205620: Here is a solution by MathLover1:
http://www.algebra.com/tutors/students/your-answer.mpl?question=1205618
I would never in a million years trust her solution, given her reputation on this forum. Can someone either verify, or provide the right solution?
You can put this solution on YOUR website! .
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 cm.
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
