A toy is in the form of a hemisphere surmounted by a right circular co

1.	A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is 2/3 of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy.(use π = \(\frac{22}7\))
Answer» Radius of base = 21cm and its volume = \(\frac{2}3\) × volume of the hemisphere ⇒ \(\frac{1}3\)πr²h = \(\frac{2}3\times\frac{2}3\)πr³ ⇒ h = \(\frac{4}3r\) = \(\frac{4}3\times21\) = 28 cm Hence surface area of the toy = Surface area of the cone + Surface area of hemisphere = π r (\(\sqrt{21^2 + 28^2}\)) + 2π r² = π (21)[35 + 2 × 21] = 5082 cm²

A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is 2/3 of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy.(use π = \(\frac{22}7\))

Answer»

Radius of base = 21cm and its volume = \(\frac{2}3\) × volume of the hemisphere

⇒ \(\frac{1}3\)πr²h = \(\frac{2}3\times\frac{2}3\)πr³

⇒ h = \(\frac{4}3r\) = \(\frac{4}3\times21\) = 28 cm

Hence

surface area of the toy = Surface area of the cone + Surface area of hemisphere

= π r (\(\sqrt{21^2 + 28^2}\)) + 2π r²

= π (21)[35 + 2 × 21]

= 5082 cm²