a conical heap formed when a farmer pores foodgrains on a ground . the slant height of base is 35 cms .the circumference of base is 132 cms . what amount of tarpolene is needed to cover grains .what is the quantity of grains with him
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Solution:-
Circumference = 132 cm
⇒ 2πr = 132
⇒ 2*22/7*r = 132
⇒ r = (132*7)/44
⇒ r = 21 cm
Radius of the base is 21 cm
Amount of Tarpolene = Curved surface area of cone = πrl
⇒ 22/7*21*35
⇒ 2310 cu cm
As we know that,
l² = √h² + r²
35² = √h² + 21²
h² = √1225 - 441
h² = 784
Height = 28 cm
Volume of food grain = 1/3πr²h
⇒ 1/3*22/7*21*21*28
⇒12936 cu cm
1 cubic centimeter of volume equals to 0.0024 kg
So, 12936 cu cm = 12936*0.0024
= 31.04 kg or 31 kg (approx)
So, the quantity of food grains with the farmer is 31 kg.
Answer.
Circumference = 132 cm
⇒ 2πr = 132
⇒ 2*22/7*r = 132
⇒ r = (132*7)/44
⇒ r = 21 cm
Radius of the base is 21 cm
Amount of Tarpolene = Curved surface area of cone = πrl
⇒ 22/7*21*35
⇒ 2310 cu cm
As we know that,
l² = √h² + r²
35² = √h² + 21²
h² = √1225 - 441
h² = 784
Height = 28 cm
Volume of food grain = 1/3πr²h
⇒ 1/3*22/7*21*21*28
⇒12936 cu cm
1 cubic centimeter of volume equals to 0.0024 kg
So, 12936 cu cm = 12936*0.0024
= 31.04 kg or 31 kg (approx)
So, the quantity of food grains with the farmer is 31 kg.
Answer.
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