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Surface Areas and Volumes Class 10 Chapter 13 Excercise: 13.5
Answer:
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
Hence the mass of wire is.
Answer:
The volume of the double cone will be = Volume of cone 1 + Volume of cone 2.
Now the surface area of a double cone is :
Answer:
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.
Answer:
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.
Answer:
From this, we can write the values of both the radius (upper and lower) and the height of the frustum.
Thus slant height of frustum is :
Now, the area of the tin shed required :
= Area of frustum + Area of the cylinder
Answer:
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.
Answer:
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
