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.

Answer 1

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Solution:

Let ORN be the cone then given, radius of the base of the cone r1 = 8 cm and height of the cone, (h) OM = 12 cm

Let P be the mid-point of OM, then OP = PM = = 6 cm

Now,

~

$\therefore\frac{OP}{OM}= \frac{PD}{MN}\\\\\implies\frac{6}{12}= \frac{PD}{8} \\\\\\implies PD = 4\:\:cm$

The plane along CD divides the cone into two parts, namely

(i) a smaller cone of radius 4 cm and height 6cm and

(ii) frustum of a cone for which  

Radius of the top of the frustum, r1 = 4 cm

Radius of the bottom,r2 = 8 cm and

height of the frustum, h = 6

$We\:\:know, \:\:Volume \:\:of \:\:cone = \frac{1}{3}\pi r^2h \\and\\Volume \:\:of \:\:frustrum = \frac{\pi h}{3} (R^2+Rr+r^2)\\\\\\\therefore\:\: Volume \:\:of \:\:cone = \frac{1}{3}\pi (4)^2(6) = 100.48\\and\\Volume \:\:of \:\:frustrum =2\pi (64+32+16)= 224\frac{22}{7} = 703.36\\\\\\\therefore\frac{V_1}{V_2} = \frac{703.36}{100.48} = \frac{7}{1}\\\\Hence,\:\: ratio \:\:of \:\:the \:\:volumes\:\:is\:\:7 :1$

Answer 2

Answer:

• ratio of the volumes of two parts is 7:1

step-by-step explaination:

Given:-

• radius of cone = 8cm.
• height = 12cm.

To Find:-

• the ratio of the volumes
• of two parts.

Solution:-

$\:\:\:\:\:$ Using concept:

(i) Volume of a cone= 1/3πr²h , 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.

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The radius of the original cone is 8cm and the height is 12.

• Then, the volume of the original cone = $\sf\frac{1}{3} \pi r^2h=\frac{1}{3} \pi (8)^2(12)=256\pi$

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 =

• $\sf\frac{12}{2}=6$

And, the radius of the small cone =

• $\sf\frac{8}{2}=4$

Volume of the small cone=

• $\sf\frac{1}{3} \pi(4)^2(6) =32\pi$

Therefore,

Volume of the frustum = Volume of the original cone - Volume of the small cone

• $\sf=256\pi -32\pi = 224\pi$

Compare the volume of the two part:

• Volume of the frustum : Volume of the small cone $\sf= 224\pi : 32\pi$

 

Volume of the frustum : Volume of the small cone =7:1

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