A right circular cone is divided …

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The height of a Cone, 3h is trisected by2 planes // to the base of the cone at equal distances. So, the cone is divided into a smaller cone & 2 frustums of the cone. The height of each piece is ‘h’ unit Since, right triangle ABG ~ tri ACF ~ tri ADE ( by AAA similarity criterion. So corresponding sides are to be proportional. So, AB/AC = h/2h = 1/2 = BG/CF = r/2r AB/AD = h/3h = 1/3 = BG/ DE = r/ 3r Now, we find the volume of each piece.. a smaller cone & 2 frustums Volume of Cone ABG = 1/3 pi r² h ………….(1) Volume of middle frustum = 1/3 pi ( r² + 4r² + 2r² ) h = 1/3 pi 7r² h ……………….…….(2) Volume of next frustum = 1/3 pi ( 4r² + 9r² + 6r²) h = 1/3 pi 19r² h …………………….(3) Now, by finding the ratio of (1),(2)&(3) we get, (1/3pi r² h) : (1/3 pi 7r² h) : (1/3 pi 19r² h) = 1:7:19