A solid toy is in the form of a hemisphere surmounted by a right circular cone Height of the cone - Maths - Surface Areas and Volumes

Question

A solid toy is in the form of a hemisphere surmounted by a right
circular cone Height of the cone is 2cm and the diameter is 4cm If
a right circular cylinder circumscribes the solid,find how much
more space it will cover

Priyanka Kedia · Accepted Answer

Given, Height of the cone = 2 cm

And, Diameter of the cone = 4 cm

So, Radius of the cone = Radius of the hemisphere = 4/2 = 2 cm

Therefore,

$V o l u m e o f t h e c o n e = \frac{1}{3} π r^{2} h = \frac{1}{3} \times π \times 2^{2} \times 2 = \frac{8}{3} π {c m}^{3}$

And,

$V o l u m e o f t h e h e m i s p h e r e = \frac{2}{3} π r^{3} = \frac{2}{3} \times π \times 2^{3} = \frac{16}{3} π {c m}^{3}$

Thus, Total volume of the solid toy = Volume of the cone + Volume of the hemisphere

= $\frac{8}{3} π + \frac{16}{3} π = 8 π {c m}^{3}$

Now, Right circular cylinder circumscribes the solid. So,

Radius of right circular cylinder CO = EF = 2 cm

And, Height of right circular cylinder OO' = OP + PO' = 2 + 2 = 4 cm

Therefore,

$V o l u m e o f r i g h t c i r c u l a r c y l i n d e r = π r^{2} h = π 2^{2} \times 4 = 16 π$

Thus,

Volume of right circular cylinder - Volume of solid toy $= 16 π - 8 π = 8 π = 8 \times 3.14 = 25.12 {c m}^{3}$

Hence, right circular cylinder is having 25.12 ${c m}^{3}$ more space than the solid.

A solid toy is in the form of a hemisphere surmounted by a right circular cone.Height of the cone is 2cm and the diameter is 4cm. If a right circular cylinder circumscribes the solid,find how much more space it will cover.