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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This answer was edited.Given perimeter of the base of a cone = 44 cm2πr = 442×22/7×r = 44r = 44×7/(2×22)r = 7 cmSlant height,l= 25height, h = √(l^{2}-r^{2})h = √(25^{2}-7^{2})h = √(625-49)h = √576h = 24 cmVolume of the cone, V = (1/3)πr^{2}hV = (1/3)×(22/7)×7^{2}×24V = (22/7)×49×8V = 22×7×8V = 1232Hence the volume of the cone is 1232 cm^{3}.Curved surface area of the cone = πrl=(22/7)×7×25= 22×25= 550 cm^{2}Hence the curved surface area of the cone is 550 cm^{2}.