It is also of interest to pupils to see at this time the method of drawing an ellipse by means of a pencil stretching a string band that moves about two pins fastened in the paper. This is a practical method, and is familiar to all teachers who have studied analytic geometry. In designing elliptic arches, however, three circular arcs are often joined, as here shown, the result being approximately an elliptic arc.
Here O is the center of arc BC, M of arc AB, and N of arc CD. Since XY is perpendicular to BM and BO, it is tangent to arcs AB and BC, so there is no abrupt turning at B, and similarly for C.[90]
Theorem. The volume of a circular cone is equal to one third the product of its base by its altitude.
It is easy to prove this for noncircular cones as well, but since they are not met commonly in practice, they may be omitted in elementary geometry. The important formula at this time is v = 1/3πr2h. As already stated, this proposition was discovered by Eudoxus of Cnidus (born ca. 407 B.C., died ca. 354 B.C.), a man who, as already stated, was born poor, but who became one of the most illustrious and most highly esteemed of all the Greeks of his time.
Theorem. The lateral area of a frustum of a cone of revolution is equal to half the sum of the circumferences of its bases multiplied by the slant height.
An interesting case for a class to notice is that in which the upper base becomes zero and the frustum becomes a cone, the proposition being still true. If the upper base is equal to the lower base, the frustum becomes a cylinder, and still the proposition remains true. The proposition thus offers an excellent illustration of the elementary Principle of Continuity.
Then follows, in most textbooks, a theorem relating to the volume of a frustum.
In the case of a cone of revolution v = (1/3)πh(r2 + r'2 + rr'). Here if r' = 0, we have v = (1/3)πr2h, the volume of a cone. If r' = r, we have v = (1/3)πh(r2 + r2 + r2) = πhr2, the volume of a cylinder.