This is the Important question of class 10 Based on Mensuration Chapter of M.L Aggarwal book for ICSE BOARD.
The volume of a right circular cone and the area of its base is given .
Find (i) the slant height of the cone. (ii) total surface area of the cone.
This is the Question Number 16, Exercise 17.2 of M.L Aggarwal.
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The volume of a right circular cone is 9856 cm cube and the area of its base is 616 cm square . Find (i) the slant height of the cone. (ii) total surface area of the cone.
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Given base area of the cone = 616 cm2
πr2 = 616
(22/7)×r2 = 616
r2 = 616×7/22
r2 = 196
r = 14
Given volume of the cone = 9856 cm3
(1/3)πr2h = 9856
(1/3)×(22/7)×142 ×h = 9856
h = (9856×3×7)/(22×142)
h = (9856×3×7)/(22×196)
h = 48
(i) Slant height, l = √(h2+r2)
l = √(482+142)
l = √(2304+196)
l = √(2500
l = 50
Hence the slant height of the cone is 50 cm.
(ii) Total surface area of the cone = πr(l+r)
= (22/7)×14×(50+14)
= 22×2×64
= 2816 cm2
Hence the total surface area of the cone is 2816 cm2.