An Important Question of class 10 Based on Mensuration Chapter of M.L Aggarwal for ICSE BOARD.
Here given the perimeter of the base of a cone and the slant height .
Find the volume and the curved surface of the cone.
This is the Question Number 15, Exercise 17.2 of M.L Aggarwal.
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The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.
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Given perimeter of the base of a cone = 44 cm
2πr = 44
2×22/7×r = 44
r = 44×7/(2×22)
r = 7 cm
Slant height, l = 25
height, h = √(l2-r2)
h = √(252-72)
h = √(625-49)
h = √576
h = 24 cm
Volume of the cone, V = (1/3)πr2h
V = (1/3)×(22/7)×72×24
V = (22/7)×49×8
V = 22×7×8
V = 1232
Hence the volume of the cone is 1232 cm3.
Curved surface area of the cone = πrl
= (22/7)×7×25
= 22×25
= 550 cm2
Hence the curved surface area of the cone is 550 cm2.