You know what a regular tetrahedron looks like.

You know what a regular tetrahedron looks like.

It's a horn shape with four equilateral triangles.

 

Generally, the volume of a three-dimensional shape is the area * height of the base.

This horn-shaped figure should be multiplied by a third.

 

To sum up

Volume of a spheroid = Area * Height / 3 of the underside.

 

The area of the underside of the surface is shown in the figure.

The area of the equilateral triangle.

A is the length of one side of an equilateral triangle.

The area of ABC is . ABC ² / 4. It's helpful to memorize this.

This is readily available only with the Pythagorean theorem.

Let's get it.

When you call E for repairs made to the BC at the extreme point A, the term E is omitted from the AEC, (lines, length. Look at the painting and take care of it.^^)

AE ² = a ² + (a / 2) ² = (3/4) a ², AE= √2

For example, the area of ABC = BC * AE2 = a * * √2 = = 3 a ² / 4

 

The next important thing is to get the height.

Take note of G ...This is the foot of the repair that came down from the base of the apparent vertex D.

You have to understand that this G dot is the center of gravity of the ABC.

 

By the nature of the center of gravity, AG : GE = 2:1.

So AG = (2/3) AE=23* √= √ 3 = = = = = a

 

Now, please watch ADG.I know the length of the two sides.

The Pythagorean theorem may be high.

I'm going straight into the bill.

H = (a ²) = (2/3 ²) = ((3) = (√ 6) a

 

The volume of the spheroid V = = a ² / 4 * (( .) / 3 = (√022) a ³. 

 

Yes. So far, the corners are generally pulled apart by a length. 

It says 2 munje high in the problem.

 H = (√3) a = 2 √ and a = 3 √ to the volumetric.

V = (√022) a ³ = (√2) ³ = 9

 

** How can I find the length of the lower part of an isosceles triangle with the length of 8? I don't ... I don't ...

There must be no height, no touch, nothing.