# Thread: volume of a prisim

1. ## volume of a prisim

good afternoon,

I have a problem. I am trying to figure out the volume of a solid, see attached image. The base is 36 SF and the top is 49SF, the height is 58ft. The base and the top are both right triangles with the base a=6 b=6 and the top a=7 b=7. Can anyone help me with a formula to figure volume? and what type of solid id this?

thanks!

2. you have to forgive me.. the base should be 18 and the top 24.5..

thank you

3. Originally Posted by vixtran
good afternoon,

I have a problem. I am trying to figure out the volume of a solid, see attached image. The base is 36 SF and the top is 49SF, the height is 58ft. The base and the top are both right triangles with the base a=6 b=6 and the top a=7 b=7. Can anyone help me with a formula to figure volume? and what type of solid id this?

thanks!
I'd call it a truncated inverted right triangular pyramid.

It is the top part of an inverted large vertical right triangular pyramid that is cut horizontally at 58 ft from the top.

The large pyramid has a total vertical height of (y +58) ft.

To solve for the y, draw an inverted triangle whose top is 7 ft wide.
Then 58 ft below the top is 6 ft wide.
The bottom is zero ft wide, of course.
By proportion,
(y +58)/7 = y/6
Cross multiply,
6(y +58) = 7y
6y +348 = 7y
348 = 7y -6y
y = 348 ft
So, the total height of the large pyramid is 348 +58 = 406 ft.

volume of pyramid is (1/3)(area of base)(altitude)

The volume, V, of the large pyramid is
V = (1/3)(24.5)(406) = 3315.67 cu.ft.

The volume, V1, of the cut pyramid is
V1 = (1/3)(18)(348) = 2088 cu.ft.

Therefore, the volume of that truncated thing is
3315.67 -2088 = 1227.67 cu.ft. -------answer.

4. ## Truncated Rectangular Pyramid

This is straight forward. It is an inverted truncated rectangular pyramid.
The formula is
Volume = 1/6((2abh+ayh+bxh+2xyh)); where a and b are the length and width of the top, x and y are the length and width of the bottom, and h is the height. This formula is very completed proving it here; but it is straight-forward.
In your case, input your values, with h = 58ft. I should assume your use of SF is square foot. Hence you can get the sides of the top and bottom rectangular shapes easily.I can post a spreadsheet I have developed for the truncated rectangular pyramid if you still having problems.
B
Originally Posted by ticbol
I'd call it a truncated inverted right triangular pyramid.

It is the top part of an inverted large vertical right triangular pyramid that is cut horizontally at 58 ft from the top.

The large pyramid has a total vertical height of (y +58) ft.

To solve for the y, draw an inverted triangle whose top is 7 ft wide.
Then 58 ft below the top is 6 ft wide.
The bottom is zero ft wide, of course.
By proportion,
(y +58)/7 = y/6
Cross multiply,
6(y +58) = 7y
6y +348 = 7y
348 = 7y -6y
y = 348 ft
So, the total height of the large pyramid is 348 +58 = 406 ft.

volume of pyramid is (1/3)(area of base)(altitude)

The volume, V, of the large pyramid is
V = (1/3)(24.5)(406) = 3315.67 cu.ft.

The volume, V1, of the cut pyramid is
V1 = (1/3)(18)(348) = 2088 cu.ft.

Therefore, the volume of that truncated thing is
3315.67 -2088 = 1227.67 cu.ft. -------answer.