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)

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Fig(2)

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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)