A heap of wheat is in the form of a cone of diameter 9m and height 3.5m.Find its volume.How much canvas is required to just cover the heap ?
Answers
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138
Solution:-
Given : Diameter = 9 m or radius = 9/2 = 4.5 m and h = 3.5 m
Volume of cone = 1/3πr²h
= 1/3*22/7*4.5*4.5*3.5
= 1559.25/21
Volume of the cone = 74.25 cu m
Just to cover the heap, we have to find the curved surface area of the cone. For slant height 'l' is needed.
So, l = √r²+h²
l = √4.5²+3.5²
l = √20.25+12.25
l = √32.5
l = 5.7008 m
Curved surface area of the cone = πrl
= 22/7*4.5*5.7008
= 564.3792/7
CSA of the cone = 80.6256 m²
So, area of the canvas required to cover the heap of wheat is 80.6256 sq m
Answer.
Given : Diameter = 9 m or radius = 9/2 = 4.5 m and h = 3.5 m
Volume of cone = 1/3πr²h
= 1/3*22/7*4.5*4.5*3.5
= 1559.25/21
Volume of the cone = 74.25 cu m
Just to cover the heap, we have to find the curved surface area of the cone. For slant height 'l' is needed.
So, l = √r²+h²
l = √4.5²+3.5²
l = √20.25+12.25
l = √32.5
l = 5.7008 m
Curved surface area of the cone = πrl
= 22/7*4.5*5.7008
= 564.3792/7
CSA of the cone = 80.6256 m²
So, area of the canvas required to cover the heap of wheat is 80.6256 sq m
Answer.
Answered by
4
Step-by-step explanation:
diameter=9m
radius=9/2m
height=3.5/10=7/2
Volume of cone= 1/3πr^2h
By subtitles value
1/3*3.14*81/4*7/2 m^2
74.1825
Let, the slant height be l
l = r^2+h^2
=81/4+49/4
l^2=130/4
l=√130/4
=114/2=5.7m
then,
Area of canvas=πrl
=3.14*9/2*57/10 m^2
=80.54m^2
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