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The radius of the base of a solid cone is 9 cm and its height is 21 cm. It is cut into 3 parts by two cuts, which are parallel to its base. The cuts are at height of 7 cm and 14 cm from the base respectively. What is the ratio of volume of top, middle and bottom parts respectively?

The radius of the base of a solid cone is 9 cm and its height is 21 cm. It is cut into 3 parts by two cuts, which are parallel to its base. The cuts are at height of 7 cm and 14 cm from the base respectively. What is the ratio of volume of top, middle and bottom parts respectively? Correct Answer 1 : 7 : 19

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The height of a Cone, 21 cm is trisected by 2 planes // to the base of the cone at the distance of 7 cm and 14 cm from the base So, the cone is divided into a smaller cone & 2 frustums of the cone. The height of each piece is ‘7’ unit

Since, right triangle ADG ~ tri AEH ~ tri AFC (by AAA similarity, corresponding sides are to be proportional.

So, AD/AE =  h/2h = 1/2 = DG/EH = r/2r

AD/AF = h/3h = 1/3 = DG/ FC = r/3r

Now, we find the volume of each piece, a smaller cone & 2 frustums

Volume of Cone ADG = 1/3 πr2 h      ---- (1)

Volume of middle frustum = 1/3 π (r2 + 4r2 + 2r2) h

= 1/3 π 7r2 h      ---- (2)

Volume of next frustum = 1/3 π (4r2 + 9r2 + 6r2) h

= 1/3 π 19r² h    ---- (3)

Now, by finding the ratio of (1), (2) & (3)

we get, (1/3π r2 h) : (1/3 π 7r2 h) : (1/3 π19r2 h)

= 1 : 7 : 19

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