Question

A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

NCERT Class X
Mathematics - Exemplar Problems

Chapter _SURFACE AREAS AND VOLUMES

Answer 1

Concept we will be using:
(i) Volume of a cone= $\frac{1}{3} \pi r^2h$ , where r=radius of the base of the cone and h is the height of the cone.

ii) Mid-point theorem: In a triangle, the line segment that joins the midpoints of the two sides of the triangle is parallel to the third side and half of it.
----------------------------------------------------------------------------------------

The radius of the original cone is 8cm and the height is 12.
Then, the volume of the original cone = $\frac{1}{3} \pi r^2h=\frac{1}{3} \pi (8)^2(12)=} 256\pi$

The cone is divided into two equal parts by drawing a plane through the mid points of its axis and parallel to the base.

Then, height of the top part (Please refer to the image) will be half of the original height.

Then, the height of the small cone = $\frac{12}{2}=6$
And, the radius of the small cone = $\frac{8}{2}=4$

Volume of the small cone=$\frac{1}{3} \pi(4)^2(6) =32\pi$

Therefore, volume of the frustum
=Volume of the original cone - Volume of the small cone
$=256\pi -32\pi = 224\pi$

Compare the volume of the two part:
Volume of the frustum : Volume of the small cone $= 224\pi : 32\pi$
 
Volume of the frustum : Volume of the small cone=7:1


Answer : Required ratio of the two parts is 7:1
 

Answer 2

Answer:

Volume of the whole cone = 1/3 x 22/7 x 8² x 12 = 804.57  cm³

Radius, r, of small cone is found by similarity:

8/12 = r/6

r = 4cm

Volume of the small cone cut off = 1/3 x 22/7 x 4² x 6 = 100 4/7

Volume of frustum = 804.57 - 100.57 = 704 cm³

The ratio of the two volumes = 100 4/7 : 704 = 1: 7

Step-by-step=