An Important Question of M.L Aggarwal book of class 10 Based on Mensuration Chapter for ICSE BOARD.

The ratio of the base radii of two right circular cones of the same height is given.

Find the ratio of their volumes

This is the Question Number 13, Exercise 17.2 of M.L Aggarwal.

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# (a) The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes. (b) The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes. (c) The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find: (i) the ratio of their volumes. (ii) the ratio of their lateral surface areas.

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(a) Let r_{1}and r_{2 }be the radius of the given cones and h be their height.Ratio of radii, r_{1}:r_{2}= 3:4Volume of cone, V_{1}= (1/3)r_{1}^{2}hVolume of cone, V_{2}= (1/3)r_{2}^{2}hV_{1}/V_{1}= (1/3)r_{1}^{2}h/ (1/3)r_{2}^{2}h= r_{1}^{2}/ r_{2}^{2}= 3^{2}/4^{2}= 9/16Hence the ratio of the volumes is 9:16.(b) Let h_{1}and h_{2 }be the heights of the given cones and r_{1}and r_{2 }be their radii.Ratio of heights, h_{1}:h_{2}= 5:2Ratio of radii, r_{1}:r_{2}= 2:5Volume of cone, V_{1}= (1/3)r_{1}^{2}h_{1}Volume of cone, V_{2}= (1/3)r_{2}^{2}h_{2}V_{1}/V_{1}= (1/3)r_{1}^{2}h_{1}/ (1/3)r_{2}^{2}h_{2}= r_{1}^{2}h_{1}/ r_{2}^{2}h_{2}= 2^{2}×5/5^{2}×2= 4×5/25×2= 20/50 = 2/5Hence the ratio of the volumes is 2:5.(c) Let r be the radius of bigger cone. Then the radius of smaller cone is r/2.Let h be the height of bigger cone. Then the height of smaller cone is h/2.(i)Volume of bigger cone, V_{1}= (1/3)r^{2}hVolume of smaller cone, V_{2}= (1/3)(r/2)^{2}(h/2) = (1/3)r^{2}h/8Ratio of volume of smaller cone to bigger cone, V_{2}/V_{1}= ( 1/3)r^{2}h/8÷ (1/3)r^{2}h= (1/24) r^{2}h ×(3/r^{2}h)= 1/8Hence the ratio of their volumes is 1:8.(ii)slant height of bigger cone = √(h^{2}+r^{2})slant height of smaller cone = √((h/2)^{2}+(r/2)^{2}) = √(h^{2}/4+r^{2}/4) = ½ √(h^{2}+r^{2})Curved surface area of bigger cone, s1 = rl=r√(h^{2}+r^{2})Curved surface area of smaller cone, s2 = rl=×(r/2)× ½ √(h^{2}+r^{2})= ¼ r√(h^{2}+r^{2})ratio of curved surface area of smaller cone to bigger cone, s2/s1 = ¼ r√(h^{2}+r^{2}) ÷ r√(h^{2}+r^{2})= ¼ r√(h^{2}+r^{2}) ×1/(r√(h^{2}+r^{2}))= 1/4Hence the ratio of curved surface area of smaller cone to bigger cone is 1:4