What is the volume of this shape, and how can I calculate it?

Question

I have this shape with $7$ vertexes as specified. You start with a unit square, then at one back edge, two vertexes go directly above the bottom two with a height of $h$. Then a special final vertex lies halfway between the previous two (in the $xy$ plane), but only half as high.

I decided to keep $h$ as a variable. But how do I determine its volume? I can't figure out a way to deconstruct it into simpler shapes such as a pyramid.

Answer 1

The shape is symmetrical about the plane $x = 0.5$. From point $(1,0,0)$, there are two pyramid that when added make one (symmetrical) half of the solid.

So the total volume will as follows

$ V = 2 \left(\dfrac{1}{3}\right) \bigg( (1)(0.5) \left(\dfrac{h + 0.5 h}{2} \right) + (0.5) (1) \left(\dfrac{ 0 + 0.5 h} { 2} \right) \bigg)$

And this simplifies to

$ V = \dfrac{h}{3} $

Answer 2

Now that i understand pyramid and tetrahedron can actually be the same thing, i was able to calculate this. But my answer is different than Hosams. (Edit: not anymore.)

I look at it as a Pyramid (Shape A) and two Tetrahedrons (Shape B).

The Pyramid has V = 1/3 bh = 1/3 * 1 * 1 = 1/3.

The Tetrahedron, for me it is easier to look at the right-most face (the blue triangle) as the base because it is the easiest triangle to calculate. That triangle has A = 1/2 b h = 1/2 * 1 * h = 1/2.

The height of that tetrahedron extends in the x direction to the special 7th vertex, so the height = 1/2. Therefore V = 1/3 b h = 1/3 * A * 1/2 = 1/12.

But there are two of those tetrahedron, so the total volume should be V = 1/3 + 2*(1/12) = 1/3 + 1/6 = 1/2. This is a different answer than Hosam Hajeer.

**Edit: ** Wait i see my error. The pyramid does not have height 1, it is actually height 1/2. Therefore it's volume is half of what i thought.

Total V = 1/6 + 2*(1/12) = 2/6 = 1/3.

Answer 3

Make a cut along the plane $z = \frac 12 y$ to create a pyramid with corners

• S = [0, 0, 0]
• T = [0, 1, 1/2]
• U = [0, 1, 1]
• V = [1/2, 1, 1/2]

The volume of a pyramid is $\frac 13 AH$ where A is the volume of a base and H is the distance from the remaining point to the plane of the base.

If you take STU as the base, then you get $V_1 = \frac 13 AH =\frac 13 \cdot (\frac 12 \cdot \frac 12 \cdot 1) \cdot \frac 12 = \frac 1 {24}$

If you take TUV as the base, then you get $V_1 = \frac 13 AH = \frac 13 \cdot (\frac 12 \cdot \frac 12 \cdot \frac 12) \cdot 1 = \frac 1 {24}$

If you remove this triangle and the mirror one, what remains is just a triangular prism, with volume $V_2 = AH = (\frac 12 \cdot \frac 12\cdot 1) \cdot 1 = \frac 14$.

So total $\frac 1{24} + \frac 1{24} + \frac 14 = \frac 13$