a right circular cone is divived into three parts by trisecting its height by two planes drawn parallel to the base. show that the volumes of the three portions starting from top are in ratio 1:7:19.
Answers
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
Answer:
Step-by-step explanaThe cone is split into three sections, call them A,B,C, from top to bottom. Section A is simply a smaller cone, and let's say it has volume x. Sections A and B combined form another cone, which has twice the dimensions of A, and therefore 8 times the volume. Then the volume of B must be 8x−x=7x. For the last part, notice that all three sections make up the original cone, which has three times the dimensions of A, and therefore 27 times the volume. Then the volume of C must be 27x−7x−x=19x.
The volumes of A,B,C are x,7x,19x, respectively, so the ratios of the three volumes are 1:7:19.tion: