Question

A solid cone of base radius 10cm is cut into two parts through mid points of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone.

Answer:

The ratio of the volume of the frustum to volume of the cone is 7:1

We are given that AQ = QP = AP/2.
In triangle AQE and triangle APC:
angle AQE = angle APC = 90 degrees
angle AEQ = angle ACP (We are given that the cone is cut parallel to the base)
Therefore, triangle AQE is similar to triangle APC and AP/AQ = 2 = AC/AE = PC/QE .. (1)
Frustum DECB:
Height = QP
R1 = PC = 10cm
From (1), PC/QE = 2 we get QE = 5 cm
R2 = QE = 5 cm
Volume = (1/3)π*QP(10² + 5² + 10*5) = (1/3)π*QP(100+25+50) = (1/3)π*175QP
Cone ADE:
Height = AQ = QP
Radius = QE = 5cm
Volume = (1/3)π*QP(5²) = (1/3)π*25QP
Therefore, the ratio of their volumes:
= (1/3)π*175QP : (1/3)π*25QP
= 7:1


Please feel free to comment below for a detailed discussion/worked out solution to this problem.