A triangle with sides 13 cm 14 cm 15 cm rotates around the middle side. What is the volume of the resulting body of revolution?

Let’s denote the vertices of this triangle A, B and C. Let AB = 15 cm, BC = 14 cm, AC = 13 cm.
The body of revolution, obtained by rotating the triangle ABC around the middle side of the BC, consists of two cones with a common base, the radius of this base r is equal to the height AD drawn to the side of rotation of the BC, forming the cones – the sides of the triangle AB and BC, the heights of the cones – segments BD and CD …
Thus, the required volume of the body is equal to the sum of the volumes of the two cones.
The volume of the cone is equal to one third of the product of the base area by the height.
V = V1 + V2 = πr ^ 2 * BD / 3 + πr ^ 2 * CD / 3 = (πr ^ 2/3) * (BD + CD) = BC * πr ^ 2/3 = BC * π * AD ^ 2/3.
Using Heron’s formula, we find the area of ​​the triangle ABC. It is equal to the root of the product of the half-perimeter of the triangle p and the differences of the half-perimeter and each of its sides: S = √p * (p-a) * (p-b) * (p-c).
p = (13 + 14 + 15) / 2 = 42/2 = 21 cm.
S = √21 * (21-13) * (21-14) * (21-15) = 84 cm2.
On the other hand, the area of ​​the triangle ABC is equal to half the product of the height AD by the side BC: S = AD * BC / 2. Hence AD ​​= 2 * S / BC = 2 * 84/14 = 12 cm.
Let’s find the required volume of the body of revolution: V = BC * π * AD ^ 2/3 = (π * 14 * 12 ^ 2) / 3 = 672π≈2111.15 cm3.