Definition D1
For a figure F in a plane α and a point V off α, the union of all segments joining V to the points of F is called the cone with base F and apex V.
Proof.
(1)The base area is the area S of F, and the height h is the distance from V to the plane α.convention
(2)If F is a polygon the cone is called a pyramid; if F is a circle, a circular cone. The proof below does not use the shape of F.convention
(3)A cone with a triangular base is the tetrahedron determined by four points, and taking any of its four faces as the base gives the same solid.convention
∎
Lemma L1
When a cone of base area S is cut by a plane parallel to the base, at distance from the apex V equal to r times the distance to the base (0 < r ≤ 1), the cross-section has area r²S.
Proof.
(1)A dilation f by r centered at V exists.A6
(2)f sends the base plane α to a plane α′ parallel to α, and the distance from V to α′ is rh.(1), A6, A7, A8, D1
(3)If P is a point of the base F, then Vf(P) = r·VP ≤ VP, so f(P) lies on segment VP, and line VP meets α′ in exactly one point, so f(P) is that point.(1), (2), A6, A12
(4)Every point of the cone lies on some VP, so f(F) is the entire intersection of the cone with α′, that is, the cross-section.(3), D1
(5)The cross-section is similar to the base with ratio r.(4), A7
(6)The similarity ratio is r, so the cross-section has area r²S.(5), A5
∎
Lemma L2
Two cones with equal base areas and equal heights have equal volumes.
Proof.
(1)Move the two cones so that both bases lie in the same plane α and both apexes lie on the same side of α. Moving them does not change the volumes.assumption, A4
(2)The heights are both h, so the two apexes lie in one plane parallel to α at distance h. Both cones lie between those two planes.(1), assumption, D1
(3)A parallel plane at distance s from α is at distance h − s from both apexes, so both cross-sections have area ((h−s)/h)²S.(2), L1, assumption
(4)The cross-sections are equal in every parallel plane between the two planes, so the two volumes are equal.(3), A1
∎
Theorem T1
A tetrahedron with base area S and height h has volume ⅓Sh.
Proof.
(1)Take a triangular prism ABC-DEF with base △ABC (area S) and height h. A↔D, B↔E, C↔F, and the lateral faces ABED, BCFE, ACFD are parallelograms.A11, convention
(2)The plane ACE divides the prism into two pieces: a tetrahedron with base △ABC and apex E, and a pyramid with base the parallelogram ACFD and apex E.(1), A9, D1
(3)A cone divides along any division of its base, so cutting ACFD by the diagonal CD into △ACD and △CFD divides the pyramid into two tetrahedra E-ACD and E-CFD.(2), D1
(4)The three tetrahedra do not overlap and together form the prism, so their volumes sum to Sh.(2), (3), A2, A3
(5)The bases △ACD and △CFD are the halves of a parallelogram cut by a diagonal, so their areas are equal; the two cones share the apex E and their bases lie in one plane, so their heights are equal too. Hence the two volumes are equal.(3), A10, L2
(6)Tetrahedron E-ABC is also the cone with base △ABE and apex C, and tetrahedron E-ACD is also the cone with base △ADE and apex C.(2), (3), D1
(7)△ABE and △ADE are the halves of parallelogram ABED cut by the diagonal AE, so their areas are equal; the apexes coincide at C and the two bases lie in one plane, so the heights are equal too. Hence the two volumes are equal.(6), (1), A10, L2
(8)All three volumes are equal, so calling the common value V, V + V + V = 3V = Sh, hence V = ⅓Sh.(4), (5), (7), A14, A15, E
(9)Tetrahedron E-ABC has base area S and height h.(1), (2), D1
(10)Any tetrahedron with base area S and height h has the same base area and height as this one, hence also volume ⅓Sh.(8), (9), L2
∎
Theorem T2
A cone with base area S and height h has volume ⅓Sh, that is, 1/3 of the volume of the cylinder with the same base area and height.
Proof.
(1)A triangle of area S exists, so a tetrahedron with that base and height h exists.A13, D1
(2)The volume of that tetrahedron is ⅓Sh.(1), T1
(3)The given cone, whether its base is a polygon or a circle, has the same base area and the same height as that tetrahedron.assumption, (1), D1
(4)By L2 the two volumes are equal, so the given cone has volume ⅓Sh.(2), (3), L2
(5)The cylinder with base area S and height h has volume Sh, so the cone's volume is 1/3 of it.(4), A2, A15
∎