a cone of radius 10 cm is divided into two parts by drawing a plane through the midpoint of its - Maths - Surface Areas and Volumes

Question

a cone of radius 10 cm is divided into two parts by drawing a
plane through the midpoint of its axis , parallel to its base
compare the volume of the two parts

Ajanta Trivedi · Accepted Answer

let the height of the cone be H and the radius be R.

this cone is divided into two parts through the mid-point of its axis. therefore AQ=1/2 AP

since $Q D ∥ P C$ therefore triangle AQD is similar to the triangle APC.

by the condition of similarity

$\frac{Q D}{P C} = \frac{A Q}{A P} = \frac{A Q}{2 A Q}$

$\frac{Q D}{R} = \frac{1}{2} \Rightarrow Q D = \frac{R}{2}$

volume of the cone ABC= $\frac{1}{3} π R^{2} H$

volume of the frustum=volume of the cone ABC-volume of the cone AED

$= \frac{1}{3} π R^{2} H - \frac{1}{3} π (\frac{R}{2})^{2} (\frac{H}{2}) = \frac{1}{3} π R^{2} H - \frac{1}{3} π \frac{R^{2} H}{8} = \frac{1}{3} π R^{2} H (1 - \frac{1}{8}) = \frac{1}{3} π R^{2} H * \frac{7}{8}$

volume of the cone AED= $\frac{1}{8} * \frac{1}{3} π R^{2} H$

therefore $\frac{v o l u m e o f t h e p a r t t a k e n o u t}{v o l u m e o f t h e r e m a i n i n g p a r t o f t h e c o n e} = \frac{\frac{1}{8} * \frac{1}{3} π R^{2} H}{\frac{7}{8} * \frac{1}{3} π R^{2} H} = \frac{1}{7}$

a cone of radius 10 cm is divided into two parts by drawing a plane through the midpoint of its axis , parallel to its base . compare the volume of the two parts