Question

A metallic bucket open at the top of height 24cm is the form of frustum of cone the radius of whose lower and upper part is 7 cm 4 cm find its volume and total surface area

Answer 1

(i). volume of frustum of cone
= ⅓ • π • h • ( R² + Rr + r² )
= ⅓ • 22/7 • 24 • ( 14² + 14 • 7 + 7² )
= 22 • 8/7 • ( 196 + 98 + 49 )
= 22 • 8/7 • 343
= 22 • 8 • 49
= 8624 cm³
(ii). slant height of frustum l
= √ ( ( R - r )² + h² )
= √ ( ( 14 - 7 )² + 24² )
= √ ( 7² + 24² )
= √ ( 49 + 576 )
= √ 625
= 25 cm
area of metallic sheet = CSA of frustum + area of base
= π • l • ( R + r ) + π • r²
= 22/7 • 25 • ( 14 + 7 ) + 22/7 • 7²
= 550/7 • 21 + 22 • 7
= 550 • 3 + 154
= 1650 + 154
= 1804 cm²

Answer 2

Mark it brainliest answer
(i). volume of frustum of cone
= ⅓ • π • h • ( R² + Rr + r² )
= ⅓ • 22/7 • 24 • ( 14² + 14 • 7 + 7² )
= 22 • 8/7 • ( 196 + 98 + 49 )
= 22 • 8/7 • 343
= 22 • 8 • 49
= 8624 cm³
(ii). slant height of frustum l
= √ ( ( R - r )² + h² )
= √ ( ( 14 - 7 )² + 24² )
= √ ( 7² + 24² )
= √ ( 49 + 576 )
= √ 625
= 25 cm
area of metallic sheet = CSA of frustum + area of base
= π • l • ( R + r ) + π • r²
= 22/7 • 25 • ( 14 + 7 ) + 22/7 • 7²
= 550/7 • 21 + 22 • 7
= 550 • 3 + 154
= 1650 + 154
= 1804 cm²