A conical heap of wheat of diameter 3.5 m is 12 m high. How much wheat is containedby the heap. About the author Everleigh
Answer: 80.54 m square Step-by-step explanation: It is given that Diameter of the conical heap =9 m Radius of the conical heap = 2 9 =4.5 m Height of the conical heap =3.5 m We know that Volume of the conical heap = 3 1 πr 2 h By substituting the values Volume of the conical heap = 3 1 ×3.14×4.5 2 ×3.5 On further calculation Volume of the conical heap =3.14×1.5×4.5×3.5 So we get Volume of the conical heap =74.1825 m 3 We know that Slant height l= (r 2 +h 2 By substituting the values l= (4.5 2 +3.5 2 ) On further calculation l= 32.5 So we get l=5.7 m We know that Curved surface area of the conical heap =πrl By substituting the values Curved surface area of the conical heap =3.14×4.5×5.7 On further calculation Curved surface area of the conical heap =80.54 m 2 Therefore, 80.54 m square of canvas is required to cover the heap of wheat. Reply
Answer:
80.54 m square
Step-by-step explanation:
It is given that
Diameter of the conical heap =9 m
Radius of the conical heap =
2
9
=4.5 m
Height of the conical heap =3.5 m
We know that
Volume of the conical heap =
3
1
πr
2
h
By substituting the values
Volume of the conical heap =
3
1
×3.14×4.5
2
×3.5
On further calculation
Volume of the conical heap =3.14×1.5×4.5×3.5
So we get
Volume of the conical heap =74.1825 m
3
We know that
Slant height l=
(r
2
+h
2
By substituting the values
l=
(4.5
2
+3.5
2
)
On further calculation
l=
32.5
So we get
l=5.7 m
We know that
Curved surface area of the conical heap =πrl
By substituting the values
Curved surface area of the conical heap =3.14×4.5×5.7
On further calculation
Curved surface area of the conical heap =80.54 m
2
Therefore, 80.54 m square
of canvas is required to cover the heap of wheat.