Question

In the adjoining figure toy total height is 20.5 CM the height of the cone which is in one end is 3.5 cm and height of the cylinder kal part is 10 cm find the volume of the toy.​

Answer 1

Correct Question:

Q. In the adjoining figure, toy total height is 20.5 cm the height of the cone which is in one end is 3.5 cm in radius and height of the cylindrical part is 10 cm find the volume of the toy.​

Given:

The total height of the toy = 20.5 cm

The height of the cylindrical part = 10 cm

The radius of the conical part = radius of the cylindrical part = r = 3.5 cm

To find:

The volume of the toy

Formula to be used:

$\boxed{\boxed{\bold{Volume\:of \:the \:cone = \frac{1}{3} \pi r^2h}}}\\\\\\\boxed{\boxed{\bold{Volume\:of \:the \:cylinder = \pi r^2h}}}$

Solution:

So,

The height of the conical part of the toy is,

= [Total height] - [Height of the cylindrical part]

= 20.5 - 10

= 10.5 cm

To get the volume of the toy, we will use the above formula to find the volume of the conical part and the volume of the cylindrical part.

∴ The volume of the conical part,

$= \frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times 10.5\\\\= \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 10.5\\\\= \frac{2829.75}{21} \\\\= \bold{134.75\:cm^3}$

∴ ∴ The volume of the cylindrical part,

$= \frac{22}{7} \times 3.5^2 \times 10\\\\= \frac{22}{7} \times 12.25 \times 10\\\\= \frac{2695}{7} \\\\= \bold{385\:cm^3}$

Now,

The total volume of the toy is,

= [Volume of conical part] + [Volume of cylindrical part]

= 134.75 cm³ + 385 cm³

= 519.75 cm³

Thus, the volume of the toy is 519.75 cm³.

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Also View:

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm.

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Base radius =3.5cm ,height=12m find the volume of cone.

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