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 => bdAreabase => acLateral edge remaining => e (from top corner to base corner)k = 1 - √(bd/ac)H= √([e/k]² - [a/2]² - [c/2]²)h = HkV = H/3*(ac-bd+bd*k)(3)Lets say Top is a rectangle with sides b & dand bottom is a rectangle with sides a & c respectively.Let height be hin that case the volume of Truncated Pyramid with rectangular base will be -V = 1/3((a²c-b²d)/(a-b))hBUT 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 HFrom parallel line property(H-h)/H = b/aRearrangingH = ah/(a-b) --------------------(1)AlsoSince V=1/3 Base area X HeightVolume of full pyramid = 1/3 X ac X HVolume 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 getV = 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)Fig(2)Lets divide the fig(1) into four different shapes as shown in fig(2)VA = Volume of cuboid = bdhVB = 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) hThen V = VA + VB + VC + VDOr, V = bdh + 1/2(a-b)dh +1/2(c-d)bh + 1/3(a-b)(c-d)hArranging and simplifying we get -V = ( h/6)(ad + bc + 2ac + 2bd)