The radius of the base of a solid cone is 9, | vedclass

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Question

(B) Let the total height of the cone be $H = 21\, cm$ and base radius be $R = 9\, cm$. The slant height $L = \sqrt{R^2 + H^2} = \sqrt{9^2 + 21^2} = \s

Vedclass · Accepted Answer

$1: 3: 5$

Vedclass · Answer

(B) Let the total height of the cone be $H = 21\, cm$ and base radius be $R = 9\, cm$. The slant height $L = \sqrt{R^2 + H^2} = \sqrt{9^2 + 21^2} = \sqrt{81 + 441} = \sqrt{522} = 3\sqrt{58}\, cm$.By similar triangles,the radius $r$ at height $h$ from the apex is given by $r/h = R/H$. The slant height $l$ at height $h$ from the apex is $l = h \cdot (L/H)$.The curved surface area of a cone is $S = \pi r l = \pi (R/H) h \cdot (L/H) h = \pi (R L / H^2) h^2$.Let the heights from the apex be $h_1, h_2, h_3$. The cuts are at $7\, cm$ and $14\, cm$ from the base,so heights from the apex are $h_1 = 21 - 14 = 7\, cm$,$h_2 = 21 - 7 = 14\, cm$,and $h_3 = 21\, cm$.The curved surface area of the top part (small cone) is $S_1 = k h_1^2 = k(7^2) = 49k$,where $k = \pi R L / H^2$.The curved surface area of the top two parts combined is $S_2 = k h_2^2 = k(14^2) = 196k$.The curved surface area of the whole cone is $S_3 = k h_3^2 = k(21^2) = 441k$.The curved surface area of the top part is $A_1 = S_1 = 49k$.The curved surface area of the middle part is $A_2 = S_2 - S_1 = 196k - 49k = 147k$.The curved surface area of the bottom part is $A_3 = S_3 - S_2 = 441k - 196k = 245k$.The ratio $A_1 : A_2 : A_3 = 49 : 147 : 245$. Dividing by $49$,we get $1 : 3 : 5$.

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