# Volume of a truncated rectangular based pyramid?

**(1)**

**Best formula to use is as follows -**

**V = h/3(Areatop + √(Areatop*Areabottom) +
Areabottom)**

**(2)**

**To find (h) using a tape measure -**

Areatop => bd

Areabase => ac

Lateral edge remaining => e (from top corner to base corner)

k = 1 - √(bd/ac)

H= √([e/k]² - [a/2]² - [c/2]²)

h = Hk

V = H/3*(ac-bd+bd*k)

(3)

Lets say Top is a rectangle with sides b & d

and bottom is a rectangle with sides a & c respectively.

Let height be h

in that case the volume of Truncated Pyramid with rectangular base will be -

**V = 1/3((a²c-b²d)/(a-b))h**

**BUT BE CAREFUL - a,b,c,d are not all independent
variables (one depends on the others) so this answer is
misleading!!!**

Proof -

Suppose the height of Full Pyramid is H

From parallel line property

(H-h)/H = b/a

Rearranging

H = ah/(a-b) --------------------(1)

Also

Since V=1/3 Base area X Height

Volume of full pyramid = 1/3 X ac X H

Volume of removed Pyramid = 1/3 X bd X (H-h)

So volume of truncated part V = 1/3(acH-bd(H-h))

=1/3((ac-bd)H + bdh)

From (1)

V = 1/3((ac-bd)ah/(a-b) + bdh)

reducing and rearranging we get

**V = 1/3((a²c-b²d)/(a-b))h**

**(4)**

**In case the truncated solid forms a prism instead, we
have following formula -**

**V = ( h/6)(ad + bc + 2ac + 2bd)**

**Proof -**

Fig(1)

<center>

</center>

Fig(2)

<center>

</center>

Lets divide the fig(1) into four different shapes as shown in fig(2)

VA = Volume of cuboid = bdh

VB = Volume of prism after joining both Bs= ½ X base X height X width = ½ (a-b) (d)(h)

VC = Volume of prism after joining both Cs = ½ X base X height X width = ½ (c-d) (b)(h)

VD = Volume of rectangular pyramid after joining all Ds = 1/3 X base area X height =

1/3 (a-b) (c-d) h

Then V = VA + VB + VC + VD

Or, V = bdh + 1/2(a-b)dh +1/2(c-d)bh + 1/3(a-b)(c-d)h

Arranging and simplifying we get -

**V = ( h/6)(ad + bc + 2ac + 2bd)**

### Why is a truncated octahedron called a truncated octahedron?

Start with an octahedron in the form of a quadrilateral based dipyramid a regular octahedron is a special case of this). This shape has six vertices - chop these off so as to remove a quadrilateral based pyramid. You will be left with a truncated octahedron. It is an octahedron which has been truncated (shortened) by cutting off its extremities.

### What is a shape that has three-dimensional shape and one rectangular face and four triangular faces?

### Shape beginning with letter t - List of geometric shapes?

Triangle Trapezoid Trapezium Tridecagon or triskaidecagon (13 sided polygon) Tetradecagon or tetrakaidecagon (14 sided polygon) Triangular prism Triangular pyramid or triangular-based Pyramid Tridecagonal pyramid or tridecagon-based pyramid Tetrahedron Torus ... and a whole host of polyhedrons of weird shapes or many sides such as Tetradecahedron (14 faced polyhedron) Triacontahedron (30 faced polyhedron) Triangular Orthobicupola (consisting of eight equilateral triangles and six squares) Triaugmented Triangular Prism (composed of 14 equilateral triangles) ... any truncated polyhedron…