# When a conical bottle rest on its flat base the water in the bottle is 8 cm from its vertex when the same conical bottle is turned upside down the water level is 2 cm from its base. What is height of the bottle ?

The key to solving this problem is the volume of water in each cone is the same. Then it becomes a matter of setting up the correct equations and simplifying using similar triangles. First let’s deal with the cone resting on its base.

**Step by Step Explanation:**

h – height of bottle

R – radius of bottle’s base

r1 – radius of circle 8 cm from vertex

The volume of water is the volume of the entire bottle minus the volume of the cone 8 cm from the vertex, which is:

(π/3)R2h – (π/3)(r1)2(8)

The two right triangles in the diagram are similar, and so we have:

r1/8 = R/h

r1 = 8R/h

We substitute this back into the volume of water formula to get:

(π/3)R2h – (π/3)(8R/h)2(8)

= (π/3)R2(h – 512/h2)

We now do a similar calculation for the other diagram.

The height and radius of the bottle is the same, so we only need to define one more variable:

r2 – radius of circle 2 cm from base

The volume of water is the volume of cone 2 cm from the base, which has a height of h – 2. So its volume is:

(π/3)(r2)2(h – 2)

Again the two right triangles in the diagram are similar, and so we have:

r2/(h – 2) = R/h

r2 = (h – 2)R/h

We substitute this back into the volume of water formula to get:

(π/3)((h – 2)R/h)2(h – 2)

= (π/3)R2(h – 2)3/h2

It seems like we have just come up with some complicated formulas. But if we keep working we will find a miraculous cancellation.

We now set the two formulas for the volume of water equal to each other.

(π/3)R2(h – 512/h) = (π/3)R2(h – 2)3/h2

The term (π/3)R2 is common to both sides of the equation, so it cancels out–it turns out the answer is independent of the radius of the bottle!

h – 512/h = (h – 2)3/h2

h3 – 512 = (h – 2)3

h3 – 512 = h3 – 6h2 + 12h – 8

6h2 – 12h – 504 = 0

h2 – 2h – 84 = 0

Now we can solve using the quadratic formula, and since height has to be positive, we only keep the solution with h > 0:

h = 1 + √85 ≈ 10.2 cm

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?

Define the variables:

h?? height of bottle

R?? radius of bottle?s base

r1?? radius of circle 8 cm from vertex

The volume of water is the volume of the entire bottle minus the volume of the cone 8 cm from the vertex, which is:

(?/3)R2h?? (?/3)(r1)2(8)

The two right triangles in the diagram are similar, and so we have:

r1/8 =?R/h

r1?= 8R/h

We substitute this back into the volume of water formula to get:

(?/3)R2h?? (?/3)(8R/h)2(8)

= (?/3)R2(h?? 512/h2)

We now do a similar calculation for the other diagram.

?

The height and radius of the bottle is the same, so we only need to define one more variable:

r2?? radius of circle 2 cm from base

The volume of water is the volume of cone 2 cm from the base, which has a height of?h?? 2. So its volume is:

(?/3)(r2)2(h?? 2)

Again the two right triangles in the diagram are similar, and so we have:

r2/(h?? 2) =?R/h

r2?= (h?? 2)R/h

We substitute this back into the volume of water formula to get:

(?/3)((h?? 2)R/h)2(h?? 2)

= (?/3)R2(h?? 2)3/h2

It seems like we have just come up with some complicated formulas. But if we keep working we will find a miraculous cancellation.

We now set the two formulas for the volume of water equal to each other.

(?/3)R2(h?? 512/h) = (?/3)R2(h?? 2)3/h2

The term (?/3)R2?is common to both sides of the equation, so it cancels out?it turns out the answer is independent of the radius of the bottle!

h?? 512/h?= (h?? 2)3/h2

h3?? 512 = (h?? 2)3

h3?? 512 =?h3?? 6h2?+ 12h?? 8

6h2?? 12h?? 504 = 0

h2?? 2h?? 84 = 0

Now we can solve using the quadratic formula, and since height has to be positive, we only keep the solution with?h?> 0:

h?= 1 + ?85 ? 10.2 cm

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