Tetrahedron

**malaygoel** · June 4th 2006, 09:26 PM

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 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.]
I think I am missing something here. Please help.

**Soroban** · June 4th 2006, 10:59 PM

Hello, malaygoel!

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.]

Place the tetrahedron on an xyz-coordinate system.

Let $A = (0,0,0),\;B = (b,0,0),\;C = (0,c,0),\;D = (0,0,d)$

The area of right triangle ABC is: $\frac{1}{2}bc$
The area of right triangle ACD is: $\frac{1}{2}cd$
The area of right triangle ABD is: $\frac{1}{2}bd$

The sum of the squares of the areas is: . $\boxed{\frac{1}{4}(b^2c^2 + c^2d^2 + b^2d^2)}$

The area of the oblique triangle BC is best found using vectors.
. . $\overrightarrow{DB} = \langle b,0,-d\rangle\qquad\qquad\overrightarrow{DC} = \langle 0,c,-d\rangle$

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

We have: $\vec{v}\;= \;\begin{vmatrix}i & j & k \\ b & 0 & -d\\ 0 & c & -d\end{vmatrix} \;= \;(cd)i + (bd)j + (bc)k$

The area of the parallelogram is: $|\vec{v}|\;=\;\sqrt{(bc)^2 + (cd)^2 + (bd)^2}$

Hence, the area of triangle ABC is: $\frac{1}{2}\sqrt{b^2c^2 + c^2d^2 + b^2d^2}$

And the square of this area is: . $\boxed{\frac{1}{4}(b^2c^2 + c^2d^2 + b^2d^2)}$ . . . Q.E.D.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I would certainly love to 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.)

**malaygoel** · June 5th 2006, 06:46 AM

Originally Posted by malaygoel

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 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.]
I think I am missing something here. Please help.

hello Soroban
It is given tatrahedron is right angled. What does it mean?

**Soroban** · June 5th 2006, 07:36 AM

Hello, malaygoel!

It is given tetrahedron is right-angled, what does it mean?

I thought it was described clearly.
There are three right angles at one vertex.
I've heard it referred to as a "trirectangular tetrahedron".

Did you sketch the "graph" I suggested?
. . $A(0,0,0),\;B(b,0,0),\;C(0,c,0),\;D(0,0,d)$

Consider a rectangular block of cheese.
Slice off one corner at an angle.
That "corner" is a right-angled tetrahedron.

**malaygoel** · June 5th 2006, 08:20 PM

Originally Posted by Soroban

Hello, malaygoel!

I thought it was described clearly.
There are three right angles at one vertex.
I've heard it referred to as a "trirectangular tetrahedron".

Did you sketch the "graph" I suggested?
. . $A(0,0,0),\;B(b,0,0),\;C(0,c,0),\;D(0,0,d)$

Consider a rectangular block of cheese.
Slice off one corner at an angle.
That "corner" is a right-angled tetrahedron.

I asm a little bit confused.
How will you define right angled tetrahedron?

**Dilanw** · October 2nd 2012, 01:55 PM

Originally Posted by malaygoel

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 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.]
I think I am missing something here. Please help.

HI Guys,

YOu can also prove this by using simple geometry but it is a bit long.

I have been obsessed with this particular theorem for nearly ten years.

Well you can clearly see this is the 3D version of the pythagoras. I am someone trying to convert everything in 2D Geometry to 3D but in a completely different way to we are being taught at universities.

In 2D we measure angles in 2D. problem is even when we try to do things in higher dimensions we try to do the same. My question to all of you is this. Suppose we start measuring the angle in the above tetrahedron with out using the usual measure. ie instead of describing it as 3 right angles in xy yz and xz planes, we measure the angle in a 3D way.

Well let me be clear

In 2D an angle is actually the arc length right?

Well in 3D it should be the surface areas on the spherical triangle.

Now in 2D one revolution is 2pie ( circumference of a circle radius 1)

in 3D it would be the surface area of the sphere radius 1. well the here we looking at the area not the arc length. well i will call it spherical area
So in 3D, 3D angles should be measured in spherical areas.

Area of a sphere radius 1 is 4pie

angle mentioned in the above mentioned tetrahedron is 4pie/8 = pie/2

so a right angle in 3D is pie/2 spherical areas.

well if anyone is interested about where im going with this get in touch

im a formar warwick uni maths graduate

**Dilanw** · October 2nd 2012, 02:00 PM

HI Guys,

YOu can also prove this by using simple geometry but it is a bit long.

I have been obsessed with this particular theorem for nearly ten years.

Well you can clearly see this is the 3D version of the pythagoras. I am someone trying to convert everything in 2D Geometry to 3D but in a completely different way to we are being taught at universities.

In 2D we measure angles in 2D. problem is even when we try to do things in higher dimensions we try to do the same. My question to all of you is this. Suppose we start measuring the angle in the above tetrahedron with out using the usual measure. ie instead of describing it as 3 right angles in xy yz and xz planes, we measure the angle in a 3D way.

Well let me be clear

In 2D an angle is actually the arc length right?

Well in 3D it should be the surface areas on the spherical triangle.

Now in 2D one revolution is 2pie ( circumference of a circle radius 1)

in 3D it would be the surface area of the sphere radius 1. well the here we looking at the area not the arc length. well i will call it spherical area
So in 3D, 3D angles should be measured in spherical areas.

Area of a sphere radius 1 is 4pie

angle mentioned in the above mentioned tetrahedron is 4pie/8 = pie/2

so a right angle in 3D is pie/2 spherical areas.

well if anyone is interested about where im going with this get in touch

im a formar warwick uni maths graduate

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