cone of radius 12 cm and height 20 cm is cut parallel to the base from the top 3 cm down and removed out the remaining part becomes the frustum of a cone find the ratio between the frustum of a cone and the cone which is removed out
Answers
Answer:
volume of cone is 2970 cm³
Step-by-step explanation:
R=12
Ht=20
volume of cone= ⅓ πr²h
⅓ 3.14×12×12×20
3017.14.
ht of smaller cone.
20/12=x/3.
x= 5cm
volume of smaller cone= 47.14 cm³
ratio = 3017.14-47.14
2970cm³
The ratio of the volume of frustum of cone by the volume of cone which is removed is equal to 63.
h1 - height of the bigger cone = 20 cm
h2 - height of the smaller cone = x
r1 - radius of the bigger cone = 12 cm
r2 - radius of the smaller cone = 3 cm
h1/h2 = r1/r2 => 20/x = 12/3
x= 5cm
The volume of the bigger cone = 1/3 × π(r1)²h1
= (22/7) × (12)² × 20
= 3014.4 cm³
The volume of cone cut off = 1/3 × π(r2)²h2
= (22/7) × (3)² × 5
= 47.1 cm³
The volume of the frustum = 3014.4 - 47.1
= 2967.3 cm³
Ratio = Volume of the frustum/Volume of the cone removed
= 2967.3/47.1 = 63