Question

the height of a right circular cone is trisected by two planes drawn parallel to the base. show that the volume of the three portions starting from the top are in the ratio 1:7:19 ​

Answer 1

Answer:see the figure labelling and calculate respectively

Step-by-step explanation:

Answer 2

Answer:

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 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

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