1. In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is
and diameter of base is 7.2 m. Find the volume of the fodder, if it is to be covered
orythin in rainy season then how much minimum polythin sheet is needed ?
(I = 22 and 117.37 = 4.17.)
Answers
Given:
- height of conical fodder=2.1 m
- diameter of conical fodder =7.2 m
- radius of conical fodder =d/2=>7.2/2=>3.6
To Find out:
Volume of fodder =?
Minimum polythin sheets required =?
Solutions:
volume of fodder = Volume of cone
Therefore,
volume of cone =1/3×pie×r²×h
putting the value
Now,
Volume of cone =1/3×22/7×3.6×3.6×2.1
=11×12.96×2.1
=28.512m²
We know that,
l²=b²+h²
l²=(3.6)²+(2.1)²
l²=12.96+4.41
l²=17.37
length=4.17 m
Now,
Length of polythin sheets =curved surface area of cone
Therefore,
Surface area of cone =pie × r ×l
putting the value
c. s.a = 22/7×3.6×4.17
= 330.264/7
= 47180.5/1000
=47.18m²
Given:
Height of conical fodder = 2.1m
Diameter of conical fodder = 7.2m
Radius of conical fodder = d/2 = 7.2/2 = 3.6m
To find out:
Volume of fodder =?
Minimum polythin sheets required =?
Solutions:
volume of fodder = volume of cone
Therefore,
volume of cone = 1/3 × πr²h
putting the value
Now,
volume of cone = 1/3 × 22/7 × 3.6 × 3.6 × 2.1
= 11 × 12.96 × 2.1
= 28.512m²
We know that:
l² = b² + h²
l² = 3.6² + 2.1²
l² = 12.96 + 4.41
l² = 17.37
l = 4.17m
Now,
Length of polythin sheets = curved surface are of cone.
Therefore,
Surface area of cone = πrl
putting the value
C.S.A = 22/7 × 3.6 × 4.17
= 330.265/7
= 47180.5/1000
= 47.18m²
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