Proposition # 14

The surface of any isosceles cone excluding the base is equal to a circle whose radius is a mean proportional
between the side of the cone ( a generator - d) and the radius ( r ) of the circle which is the base of the cone

Instead of using the proof from the book of Archimedes, I use a different version which is much shorter.
Let S be the surface of the cone excluding the base.
Let B be the surface of the base of the cone.
Let O be the center of the base
Let T be the apex of the cone
Let A be a point on the base circle
Let OA = r and TA = d
 TO is perpendicular to the base circle at O ===>  angle TAO is the angle between the surrounding surface of the cone
                                                                             and the base circle
                                                                 ===>  B = S * cos (TAO)
 Since cos(TAO) = OA / TA = r / d.    ===>  B = S * (r / d) or S = B * ( d / r)  = pi * ( r * r) * ( d / r) = pi * d * r
          or  S = pi * d * r.

So S is equal to the area of a circle whose radius is the square root of d*r

Proposition # 16

If an isosceles cone be cut by a plane parallel to the base, the portion of the surface of the cone between the parallel planes is
equal to a circle whose radius is a mean proportional between (1) the portion of the side of the cone intercepted by the parallel
planes and (2) the line which is equal to the sum of the radii of the circles in the parallel planes

Let OAB be a triangle through the axis of a cone, DE its intersection with the plane cutting off the frustum, and OFC the axis of the
cone. Then the surface of the cone OAB is equal to a circle whose radius is equal to square root of (OA* AC) ( proposition 14)
Similarly, the surface of the cone ODE is equal to a circle whose radius is equal to the square root of (OD * DF)
The surface of the frustum is equal to the difference between the two circles
Now:
OA.AC - OD.DF =  (OD + DA) AC - OD. DF = OD. AC + DA. AC - OD. DF
Since two triangle ODF and OAC are similar
OD. AC = OA.DF and OA / AC = OD / DF
Hence
OA. AC - OD.DF = DA.AC + DA.DF = DA (AC + DF)
And since circles are to one another as the squares of their radii, it follows that the difference between the circles whose radii
are sqrt (OA.AC), sqrt ( OD.DF) respectively is equal to a circle whose radius is sqrt (DA ( AC + DF))
Therefore, the surface of the frustum is equal to this circle
 
 

Proposition # 21

A regular polygon of an even number of sides being inscribed in a circle, as ABC...A'...C'B'A, so that AA' is a diameter
if two angular points next but one to each other, as B, B', be joined, and the other lines parallel to BB' and joining pairs of
angular points be drawn, as CC', DD'..., then

                  (BB' + CC' + .....): AA' = A'B: BA
Let BB', CC', DD' .... meet AA' in F, G, H, ...; and let CB', DC',.... be joined meeting AA' in K, L .... respectively.

Then CB', DC',.... are parallel to one another and to AB.

Prove CB', DC' parallel to each other and to AB as followings:
Since arc B'C' = arc C'D' = ( 30 degree)  ==> angle KCG = angle GCL  (1)
Since AA' is diameter of the circle ==> AA' _|_ CC' at G (2)
From (1) and (2) ==> two triangles KCG and GCL are equal
                          ==> angle CKG =  angle CLG (3)

Two right triangles LGC' and CGL are equal because:
                      CG = GC'
                      GL common
==> angle CLG  = angle C'LG (4)
From (3) and (4)  ==> angle CKG = angle GLC' ==> CK // LC' ==> CB' // DC' (*)
Similarly,  we will have AB // CB' // DC'

Hence, by similar triangles:
            BF / FA     =          B'F /  FK
                             =          CG / GK
                             =          G'G / GL
                                ......................................
                             =          E' I /  IA'
and summing the antecedents and consequence respectively, we have:

                (BB' + CC' +  ....) / AA'  = BF / FA

Two right triangles AFB and ABA' is similar because they have the same angle BAF ( = BAA')
so that BF / FA = A'B / BA

==>  (BB' + CC' +  ....) / AA'  = BF / FA = A'B /  BA

Proposition # 22

If a polygon be inscribed in a segment of a circle LAL' so that all its sides excluding the base are equal and their number even,
as LK.....A.....K'L', A being the middle point of the segment, and if the lines BB', CC', parallel to the base LL' and joining pairs
of angular points be drawn, then
     (BB' + CC' + ......+ LM) : AM = A'B : BA
where M is the middle point of LL' and AA' is the diameter through M

Joining CB', DC', KD', LK' as in the last proposition, and supposing that they meet AM in P,Q,J, R
while BB', CC',DD', KK' meet AM in F, G, O, H we have, by similar triangles,

                BF / FA  = B'F / FP
                            = CG / PG
                            = C'G / GQ
                            = DO / OQ
                            = D'O / OJ
                           .= KH / HJ
                           =  K'H / HR
                            = LM / RM

and summing the antecedents as consequence, we obtain
        ( BB' + CC' + .....+ LM) / AM = BF / FA = A'B / BA ( using result from proposition # 21)
 

Proposition # 23

Take a great circle ABC... of a sphere, inscribe in it a regular polygon whose sides are a multiple of four in number.
Let AA', MM' be diameters at right angles and joining opposite angular points of the polygon. Then if the polygon and
great circle revolve together about the diameter AA', the angular points of the polygon, except A, A', will describe circles
on the surface of the sphere at right angles to the diameter AA'. Also the sides of the polygon will describe portions of
conical surfaces, e.g. BC will describe a surface forming part of a cone whose base is a circle about CC' as diameter
and whose apex is the point in which CB, C'B' produced meet each other and the diameter AA'.

Comparing the hemisphere MAM' and that half of the figure described by the revolution of the polygon which is included in the
hemisphere, we see that the surface of the hemisphere and the surface of the inscribed figure have the same boundaries in one
plane (viz. the circle on MM' as diameter), the former surface entirely includes the latter, and they are both concave in the same
direction.

In other words, according to Euclidean geometry, the shortest distance between two points is the line segment goes through these
two points. So, for instance, the arc length connects B and C is longer than the line segment BC. When arc length BC rotate around
AA', it will form a part the sphere that is limited by two circles whose diameters are BB' and CC'. Meanwhile, the line segment BC
will form a part of the cone that has the same bases as the partial sphere above. Since the arc length BC is longer than the line
segment BC, it will create a larger area when they both rotate around AA'.

Therefore the surface of the hemisphere is greater than that of the inscribed figure,; and the same is true of the other
halves of the figures. Hence the surface of the sphere is greater than the surface described by the revolution of the polygon
inscribed in the great circle about the diameter of the great circle.

Proposition # 24

If a regular polygon AB...A'...B'A, the number of whose sides is a multiple of four, be inscribed in a great circle of a sphere, and
if BB' subtending two sides be joined, and all the other lines parallel to BB' and joining pairs of angular points be drawn, then
the surface of the figures inscribed in the sphere by the revolution of the polygon about the diameter AA' is equal to a circle the
square of whose radius is equal to the rectangle
                                                 BA (BB' + CC' + ...).
The surface of the figure is made up of the surfaces of parts of different cones

Now the surface of the cone ABB' is equal to a circle whose radius is sqrt(BA. 1/2 BB') (Proposition # 16)

The surface of the frustum BB'C'C is equal to a circle of radius sqrt (BC . 1/2 (BB' + CC')) and so on

It follows, since BA = BC = .. that the whole surface is equal to a circle whose radius is equal to

        sqrt( BA (BB' + CC'+ .....+ MM' + ... + YY'))
 

Proposition # 25

The surface of the figure inscribed in a sphere as in the last propositions, consisting of portions of conical surfaces, is less
than four times the greatest circle in the sphere.

Let AB...A'...B'A be a regular polygon inscribed in a great circle, the number of its sides being a multiple of four

 As before, let BB' be drawn subtending two sides, and CC',...,YY' parallel to BB'.
Let R be a circle such that the square of its radius is equal to

            AB ( BB' + CC'+ ....+ YY').
so that the surface of the figure inscribed in the sphere is equal to R (Proposition #24)
Now
        (BB' + CC'+ .....+ YY') / AA' = A'B / AB  (Proposition # 21)
whence
      AB ( BB' + CC'+ ....+ YY') = AA'. A'B
Hence      (radius of R)(radius of R) = AA'.A'B  < AA'.AA'

==> the surface of the inscribed figure ( or circle R) < pi (2. AA'/2) (2. AA'/2) = 4 circle AMA'M'

Therefore the surface of the inscribed figure, or the circle R, is less than four times of the circle AMA'M'

Proposition # 26

The figure inscribed as above in a sphere is equal in volume to a cone whose base is a circle equal to the surface of the
figure inscribed in the sphere and whose height is equal to the perpendicular drawn from the center of the sphere to one
side the polygon
 

Suppose, as before, that AB....A'...B'A is the regular polygon inscribed in a great circle, and let BB', CC', ... be joined.
With apex O construct cones whose bases are the circles on BB', CC', .... as diameters in planes perpendicular to AA'.
Then OBAB' is a solid rhombus, and its volume is equal to a cone whose base is equal to the surface of the cone ABB'
and whose height is equal to the perpendicular from O on AB. Let the length of the perpendicular be p. [Proposition 18]

   Volume of OBAB' = Surface of the cone ABB' * p (1)

   Again, if CB, C'B' produced meet in T, the portion of the solid figure which is described by the revolution of the triangle
BOC about AA' is equal to the difference between the rhombi OCTC' and OBTB', i.e. to a cone whose base is equal to
the surface of the frustum BB'C'C and whose height is p. [Proposition 20]

Volume created by rotating OCB around AA' = surface of frustum BB'C'C * p (2)

Proceeding in this manner we have:

Volume created by rotating OCM around AA' = surface of frustum CC'M'M * p (3)
Volume created by rotating OMN around AA' = surface of frustum NN'M'M * p (4)
Volume created by rotating ONY around AA' = surface of frustum NN'YY' * p   (5 )
Volume of the rhombus OYA'Y'   = surface of the cone A'YY' * p (6)

Adding up (1),(2),(3),(4),(5),(6) we have:

The volume of the solid of revolution is equal to p * (surface of cone ABB' + surface of CC'MM' +.....+ surface of cone A'YY')
or  equal to  p * the surface of the solid created by rotating the polygon around AA' or equal to a cone whose base is equal to
the surface of the solid and whose height is p.

Proposition # 27

The figure inscribed in the sphere as before is less than four times the cone whose base is equal to a great circle of the sphere
and whose height is equal to the radius of the sphere.

By Prop. 26 the volume of the solid figure is equal to a cone whose base is equal to the surface of the solid and whose height is p,
the perpendicular from O on any side of the polygon. Let R be such a cone.
Take also a cone S with base equal to the great circle, and height equal to the radius, of the sphere.
Now, since the surface of the inscribed solid is less than four times the great circle (Prop 25), the base of the cone is less than four
times the base of the cone S.
Also the height (p) of R is less than the height of S.
Therefore the volume of R is less than four times that of S; and the proposition is proved

Proposition # 28

Let a regular polygon, whose sides are a multiple of four in number, be circumscribed about a great circle of a given sphere, as
AB...A'...B'A; and about the polygon describe another circle, which will therefore have the same center as the great circle of
the sphere. Let AA' bisect the polygon and cut the sphere in a, a'.
If the great circle and the circumscribed polygon revolve together about AA', the great circle will describe the surface of a sphere
the angular points of the polygon except A, A' will move around the surface of a larger sphere, the points of contact of the sides
of the polygon with the great circle of the inner sphere will describe circles on that sphere in planes perpendicular to AA', and
the sides of the polygon themselves will describe portions of conical surfaces. The circumscribed figure will thus be greater than
the sphere itself.

Let any side, as BM, touch the inner circle in K, and let K' be the point of contact of the circle with B'M'. Then the circle described
by the revolution of KK' about AA' is the boundary in one plane of two surfaces.

(1) the surface formed by the revolution of the circular segment KaK', and (2) the surface formed by the revolution of the part
KB...A...B'K' of the polygon.
Now the second surface entirely includes the first and they are both concave in the same direction;
therefore the second surface is greater than the first.
The same is true of the portion of the surface on the opposite side of the circle on KK' as diameter.
Hence, adding, we see that the surface of the figure circumscribed to the given sphere is greater than
that of the sphere itself.

Proposition # 29

In a figure circumscribed to a sphere in the manner shown in the previous proposition the surface is equal to a circle the square
on whose radius is equal to AB ( BB' + CC' + .........).
For the figure circumscribed to the sphere is inscribed in a larger sphere, and the proof of Prop. 24 applies

Proposition # 30

The surface of a figure circumscribed as before about a sphere is greater than four times the great circle of the sphere.

Let AB....A'...B'A be the regular polygon of 4n sides which by its revolution about AA' describes the figure circumscribing
the sphere of which ama'm' is a great circle. Suppose aa', AA' to be in one straight line.
Let R be a circle equal to the surface of the circumscribed solid.
Now (BB' + CC' + ...) : AA' = A'B : BA (as in Prop. 21)
so that  AB ( BB' + CC' + ....) = AA'.A'B
Hence  ( radius of R ) = sqrt ( AA'. A'B )  (Prop. 29)
                                  > A'B

But A'B = 2 OP, where P is the point in which AB touches the circle ama'm'.

Therefore (radius of R) > (diameter of circle ama'm');
whence R, and therefore the surface of the circumscribed solid, is greater than four times the great circle of the given sphere.

Proposition # 31
 

The solid of revolution circumscribed as before about a sphere is equal to a cone whose base is equal to the surface of the solid
and whose height is equal to the radius of the sphere.
The solid is, as before, a solid inscribed in a larger sphere; and since the perpendicular on any side of the revolving polygon is equal
to the radius of the inner sphere, the proposition is identical with Prop. 26

COR: The solid circumscribed about the smaller sphere is greater than four times the cone whose base is a greater circle of the sphere
and whose height is equal to the radius of the sphere

For, since the surface of the solid is greater than four times the greater circle of the inner sphere ( Prop 30 ), the cone whose base is
equal to the surface of the solid and whose height is the radius of the sphere is greater than four times the cone of the same height which
has the great circle for base (lemma 1)
Hence, by the proposition, the volume of the solid is greater than four times the latter cone.