A solid toy is in the form of a hemisphere
surmounted by a right circular cone
Height of the cone is 4 cm and the
diameter of the base is 8 cm. If a right
circular cylinder circumscribes the solid.
Find how much more space it will cover?
Answers
Given:
✰ A solid toy is in the form of a hemisphere surmounted by a right circular cone.
✰ Height of the cone = 4 cm
✰ Diameter of the base of the cone = 8 cm
✰ A right circular cylinder circumscribes the solid.
To find:
✠ How much more space it will cover?
Solution:
Let QRS be the solid toy is in the form of a hemisphere surmounted by a right circular cone PQR and ABCD be the right circular cylinder which circumscribes the solid.
Refer the attachment
then,
Radius of the base of the cone = Radius of the hemisphere = Radius of the base of the cylinder = 4 cm
⇾ ER = ES = SC = 4 cm
Now,
⟹ The height of the cone = 4 cm
⟹ The height of the cylinder = PE + ES
⟹ The height of the cylinder = 4 + 4
⟹ The height of the cylinder = 8 cm
First we will find out the volume of the cylinder and then the volume of a solid by using formula. Putting the values in and then doing the required calculations. After that we will subtract the volume of the solid from the volume of a cylinder to find out the space required or the space it will cover.
✭ Volume = πr²h ✭
➛ Volume of the cylinder = πr²h
➛ Volume of the cylinder = π × 4² × 8
➛ Volume of the cylinder = π × 16 × 8
➛ Volume of the cylinder = ( π × 128 ) cm³
➛ Volume of the solid toy = Volume of cone + Volume of hemisphere
➛ Volume of the solid toy = 1/3 πr²h + 2/3πr³
➛ Volume of the solid toy = 1/3π × ( 4 )² × 4 + 2/3π × (4)³
➛ Volume of the solid toy = 1/3π × 16 × 4 + 2/3π × 64
➛ Volume of the solid toy = ( 1/3π × 64 + 2/3π × 64 ) cm³
➤ Required space = Volume of the cylinder - Volume of the solid toy
➤ Required space = ( π × 128 ) - ( 1/3π × 64 + 2/3π × 64 )
➤ Required space = ( π × 128 ) - ( 64 × π )
➤ Required space = 128π - 64π
➤ Required space = 64π cm³
Now, if we multiply it by π
➤ Required space = ( 64 × 22/7 ) cm³
➤ Required space ≈ 201.14 cm³
∴ It will cover 64π cm³ more space.
_______________________________
Answer:
Given:
✰ A solid toy is in the form of a hemisphere surmounted by a right circular cone.
✰ Height of the cone = 4 cm
✰ Diameter of the base of the cone = 8 cm
✰ A right circular cylinder circumscribes the solid.
To find:
✠ How much more space it will cover?
Solution:
Let QRS be the solid toy is in the form of a hemisphere surmounted by a right circular cone PQR and ABCD be the right circular cylinder which circumscribes the solid.
Refer the attachment
then,
Radius of the base of the cone = Radius of the hemisphere = Radius of the base of the cylinder = 4 cm
⇾ ER = ES = SC = 4 cm
Now,
⟹ The height of the cone = 4 cm
⟹ The height of the cylinder = PE + ES
⟹ The height of the cylinder = 4 + 4
⟹ The height of the cylinder = 8 cm
First we will find out the volume of the cylinder and then the volume of a solid by using formula. Putting the values in and then doing the required calculations. After that we will subtract the volume of the solid from the volume of a cylinder to find out the space required or the space it will cover.
✭ Volume = πr²h ✭
➛ Volume of the cylinder = πr²h
➛ Volume of the cylinder = π × 4² × 8
➛ Volume of the cylinder = π × 16 × 8
➛ Volume of the cylinder = ( π × 128 ) cm³
➛ Volume of the solid toy = Volume of cone + Volume of hemisphere
➛ Volume of the solid toy = 1/3 πr²h + 2/3πr³
➛ Volume of the solid toy = 1/3π × ( 4 )² × 4 + 2/3π × (4)³
➛ Volume of the solid toy = 1/3π × 16 × 4 + 2/3π × 64
➛ Volume of the solid toy = ( 1/3π × 64 + 2/3π × 64 ) cm³
➤ Required space = Volume of the cylinder - Volume of the solid toy
➤ Required space = ( π × 128 ) - ( 1/3π × 64 + 2/3π × 64 )
➤ Required space = ( π × 128 ) - ( 64 × π )
➤ Required space = 128π - 64π
➤ Required space = 64π cm³
Now, if we multiply it by π
➤ Required space = ( 64 × 22/7 ) cm³
➤ Required space ≈ 201.14 cm³
∴ It will cover 64π cm³ more space.
_______________________________
Step-by-step explanation:
Hope this answer will help you.✌️