Question

derive the formula of volume of frustum

Answer 1

V1=13A1y

V2=13A2(y−h)
 

V=V1−V2=13A1y−13A2(y−h)

V=13A1y−13A2y+13A2h

V=13[(A1−A2)y+A2h]   →   Equation (1)
 

By similar solids (Click here for more information about Similar Solids):
A2A1=(y−hy)2

√A2A1=1−hy

hy=1−√A2A1=1−√A2√A1

hy=√A1−√A2√A1

yh=√A1√A1−√A2

y=√A1√A1−√A2h=(√A1√A1−√A2×√A1+√A2√A1+√A2)h

y=A1+√A1A2A1−A2h
 

Substitute y to Equation (1),
V=13[(A1−A2)(A1+√A1A2A1−A2h)+A2h]

V=13[(A1+√A1A2)h+A2h]

V=13[A1+√A1A2+A2]h

V=h3[A1+A2+√A1A2]

Answer 2

The formula for frustum of a pyramid or frustum of a cone is given by
 

V=h3[A1+A2+A1A2−−−−−√]V=h3[A1+A2+A1A2]

 

Where:
h = perpendicular distance between A1 and A2 (h is called the altitude of the frustum)
A1 = area of the lower base
A2 = area of the upper base
Note that A1 and A2 are parallel to each other.
 

Derivation:
V1=13A1yV1=13A1y

V2=13A2(y−h)V2=13A2(y−h)
 

V=V1−V2=13A1y−13A2(y−h)V=V1−V2=13A1y−13A2(y−h)

V=13A1y−13A2y+13A2hV=13A1y−13A2y+13A2h

V=13[(A1−A2)y+A2h]V=13[(A1−A2)y+A2h]   →   Equation (1)
 

By similar solids (Click here for more information about Similar Solids):
A2A1=(y−hy)2A2A1=(y−hy)2

A2A1−−−√=1−hyA2A1=1−hy

hy=1−A2A1−−−√=1−A2−−−√A1−−−√hy=1−A2A1=1−A2A1

hy=A1−−−√−A2−−−√A1−−−√hy=A1−A2A1

yh=A1−−−√A1−−−√−A2−−−√yh=A1A1−A2

y=A1−−−√A1−−−√−A2−−−√h=(A1−−−√A1−−−√−A2−−−√×A1−−−√+A2−−−√A1−−−√+A2−−−√)hy=A1A1−A2h=(A1A1−A2×A1+A2A1+A2)h

y=A1+A1A2−−−−−√A1−A2hy=A1+A1A2A1−A2h
 

Substitute y to Equation (1),
V=13[(A1−A2)(A1+A1A2−−−−−√A1−A2h)+A2h]V=13[(A1−A2)(A1+A1A2A1−A2h)+A2h]

V=13[(A1+A1A2−−−−−√)h+A2h]V=13[(A1+A1A2)h+A2h]

V=13[A1+A1A2−−−−−√+A2]hV=13[A1+A1A2+A2]h

V=h3[A1+A2+A1A2−−−−−√]