It reminds me of the times of my childhood when I was playing games such as Path of Exile (PoE). I played with two friends, Jakub and Mateusz. I did not have too much money myself and at first I watched them play. But in the end I worked hard with my hard work and I could buy exalted orbs.

This is the currency that I really like and that I need. I liked them both, they taught me to play PoE and they gave me my first orbs, then I did not know what currency was for.

Before I got the first exalted orb, I learned to use poe orbs.

Today, he buys them all, it is a fast and safe way to increase his stock Exalted orbs

buy Exalted orbs]]>

Let the slant height

and the radius of the circleLet

since and , Since . From draw a perpendicular to base at point , then is a right triangle, where and . Since the radius of the smaller base we choose the volume of the frustum]]>

]]>

How do you know that JD=x?

From a point you can make two tangents to a circle. The lengths of these two tangents (from point to circle) are equal.

RandomPieKevin wrote:

How did you get the base as 20-2x?

Using the same rule AJ = AH = 10 - x so the base is AB = 20 - 2x

In triangle DAM, DM = FH = 6 and DA = 10, therefore AM = 8.

AH = 10 - x and also x + 8 => 10 - x = x + 8 => x = 1

Therefore AH = 9.

So the large cone is 9x as big as the small cone.

If the height of the small cone = h, then 9 times small height = large height => 9h = 6 + h => h = 3/4

So small cone

and large cone

Bob

]]>Bob

]]>How did you get the base as 20-2x?

____________________________________________________________________________________________________________________________________How do you know that JD=x?]]>

I don't know which base you guys are talking about. Also, is x the radius of the inscribed circle?

Don't forget this is a frustum. That is, you start with a big cone, chop off a small cone at the top, and the frustum is what remains. In post 2 I have made a diagram that shows a vertical cross section through the middle of the solids. The red outline is the frustum. It has a circular top radius x and a circular base, radius to be found.

The sphere fits inside the frustum and touches the top and bottom circles and also the sides. The sphere appears in the diagram as a circle because the cross section cuts through the middle of the sphere. It has a radius of 3.

The sloping side of the frustum is 10.

If a pair of tangents are drawn from a point to a circle they will have equal length. To prove this draw a line from the point to the centre of the circle, making two triangles. It is fairly straight forward to show that these triangles are congruent.

So the top radius, x, is repeated down the sloping side, and hence the lower distance is 10-x. By the same tangent length rule this distance is repeated across the bottom. So the base radius is 10-x.

championmathgirl's method is to drop a perpendicular from the top radius down to the bottom making a right angled triangle. It has sides of 10, 10-x-x, and 6. You can use Pythagoras to work out x.

I like to work out the volume of a frustum by calculating the volumes of the large and small cones, and then subtracting.

Bob

]]>I understand how you got your equations, but haven't worked out yet how stefy arrived at his solutions for x and y

Here is one way:

```
FindInstance[
y/(10 + y) == x/(10 - x) == Sqrt[-x^2 + y^2]/(
6 + Sqrt[-x^2 + y^2]), {x, y}, Reals, 5]
```

And then a tiny ansatz is required.

]]>I used a simpler algebra way to find x. Because you could use Pythagorean Theorem and write the base as 16+2x. Then since we also have the base as 20-2x. Setting them equal to each other we get x=1. Then use the ratios y/(y+10)=x(10-x) to find y.

I don't know which base you guys are talking about. Also, is x the radius of the inscribed circle?

]]>Bob

]]>In the image I added DM to create right-angled triangle DMA (for which the lengths of DA and DM are known) in order to obtain the length of AM (by Pythagoras).

]]>