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The required surface area is given by :
Surface area of article = Surface area of cylindrical part + 2 (Surface area of hemisphere)
Now, area of cylinder
or
or
And the surface area of hemisphere :
or
or ...

Similar to how we find the surface area of the frustum.
The volume of the frustum is given by-
=Volume of the bigger cone - Volume of the smaller cone

In the case of the frustum, we can consider:- removing a smaller cone (upper part) from a larger cone.
So the CSA of frustum becomes:- CSA of bigger cone - CSA of the smaller cone
And the total surface area of the frustum is = CSA of frustum + Area of upper circle and area of lower circle.
...

From this, we can write the values of both the radius (upper and lower) and height of frustum.
Thus slant height of frustum is :
Now, the area of the tin shed required :
= Area of frustum + Area of the cylinder

Firstly we will calculate the volume of rainfall :
Volume of rainfall :
And the volume of the three rivers is :
It can be seen that both volumes are approximately equal to each other.

The total volume of the cistern is :
And the volume to be filled in it is
Now let the number of bricks be n.
Then the volume of bricks :
Further, it is given that brick absorbs one-seventeenth of its own volume of water.
Thus water absorbed :
Hence we write :
Thus the total number of bricks is 1792.

The volume of the double cone will be = Volume of cone 1 + Volume of cone 2.
(Note that sum of heights of both the cone is 5 cm - hypotenuse).
Now the surface area of a double cone is :
...

A number of rounds are calculated by :
Thus the length of wire in 40 rounds will be
And the volume of wire is: Area of cross-section Length of wire
Hence the mass of wire is.
...

The figure for the problem is shown below :
Using geometry we can write :
and
Thus the volume of the frustum is given by :
Now, the radius of the wire is :
...

Firstly we will calculate the slant height of the cone :
Now, the volume of the frustum is :
= Capacity of the container.
Now, the cost of 1-litre milk is Rs. 20.
Then the cost of 10.449-litre milk will be
The metal sheet required for the container is :
Thus...

The area of material used is given by :
Area of material = Curved surface area of a frustum of cone + Area of upper end

We are given the perimeter of upper and lower ends thus we can find r1 and r2.
And,
Thus curved surface area of the frustum is given by :

The capacity of glass is the same as the volume of glass.
Thus the volume of glass :
= Capacity of glass

Area of the cross-section of pipe is
Speed of water is given to be = 3 km/hr
Thus, the volume of water flowing through a pipe in 1 min. is
Now let us assume that tank will be completely filled after t minutes.
Then we write :
Hence time required for filling of the tank completely is 100 minutes.

Speed of water is: 10 Km/hr
And the volume of water flow in 1 minute is :
Thus the volume of water flow in 30 minutes will be :
Let us assume irrigated area be A. Now we can equation the expression of volumes as the volume will remain the same.
Thus the irrigated area is . ...

According to question volume will remain constant thus we can write :
The volume of bucket = Volume of heap formed.
Let the radius of heap be r.
And thus the slant height will be
Hence the radius of heap made is 36 cm and its slant height is .

Let us assume the number of coins that need to be melted be n.
Then we can write :
The volume of n coins = Volume of cuboid formed.
Thus the required number of coins is 400.

Let the number of cones that can be filled with ice cream be n.
Then we can write :
The volume of a cylinder containing ice cream = n ( volume of 1 ice cream cone )
Hence the number of cones that can be filled is 10.

According to the question, the volume is conserved here :
The volume of soil dug out = Volume of the embankment made.
Let the height of the embankment is h.
Hence the height of the embankment made is 1.125 m.

According to the question, the volume of soil dug will be equal to the volume of the platform created.
Thus we can write :
The volume of soil dug = Volume of platform
Thus the height of the platform created is 2.5 m.

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