Hello, malaygoel!

Place the tetrahedron on an xyz-coordinate system.A right angled tetrahedron ABCD is one in which the three angles at the vertex A are all right angles.

Prove that the square of the area of the triangle BCD is equal to

the sums of the squares of the areas on the other three sides.

[This can be done mentally in 10 seconds with the right approach.]

Let

The area of right triangle ABC is:

The area of right triangle ACD is:

The area of right triangle ABD is:

The sum of the squares of the areas is: .

The area of the oblique triangle BC is best found using vectors.

. .

The magnitude of the cross product is the area of theparallelogramdetermined by the two vectors.

We have:

The area of the parallelogram is:

Hence, the area of triangle ABC is:

And the square of this area is: . . . . Q.E.D.

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I would certainlyloveto see that 10-second approach.

This problem was given by my Calculus 3 professor.

He asked us to find the 3-dimensional analogy to the Pythagorean Theorem.

I came up with this analogy and worked out the math in only a weekend.

(I used Heron's Formula for the oblique triangle. you see.)