A cone 7dm high and 6 dm in diameter is placed on the ground. Sand is poured over it until conical heap 21 dm high and 33 dm circumference at the bottom is formed. Find how many m^3 of sand is there?

what about 33dm circumfernce

what about 33dm circumfernce

well, you know that C = 2pi r, so,

r = 33/2pi

Forgot about that?

To solve this problem, we need to calculate the volume of the cone and the volume of the heap of sand, and then find the difference between the two volumes.

First, let's calculate the volume of the original cone. The formula for the volume of a cone is V = (1/3) * π * r^2 * h, where r is the radius and h is the height.

Given that the diameter is 6 dm, the radius (r) is therefore 6 dm / 2 = 3 dm. The height (h) of the original cone is given as 7 dm.

Using these values, we can calculate the volume of the original cone:

V_cone = (1/3) * π * (3 dm)^2 * 7 dm
= (1/3) * 3.14 * 9 dm^2 * 7 dm
= 3.14 * 9 dm^3 * 7 dm / 3
= 3.14 * 63 dm^4 / 3
≈ 66.29 dm^3

Next, let's calculate the volume of the heap of sand. The heap is also in the shape of a cone, but its dimensions are different. We are given that the heap is 21 dm high and has a circumference of 33 dm at the bottom.

To calculate the radius (r) of the heap, we can use the formula for the circumference of a circle, which is C = 2πr. Rearranging this formula, we can solve for r: r = C / (2π)

Given that the circumference is 33 dm, we have:

r_heap = 33 dm / (2π)
≈ 33 dm / 6.28
≈ 5.26 dm

Now we can calculate the volume of the heap of sand:

V_heap = (1/3) * π * (5.26 dm)^2 * 21 dm
≈ (1/3) * 3.14 * 27.61 dm^2 * 21 dm
≈ 3.14 * 27.61 dm^3 * 7 dm
≈ 575.16 dm^3

Finally, to find the volume of sand present, we subtract the volume of the original cone from the volume of the heap of sand:

V_sand = V_heap - V_cone
≈ 575.16 dm^3 - 66.29 dm^3
≈ 508.87 dm^3

However, the question asks for the volume in m^3. Since 1 m = 10 dm, we need to convert the volume from dm^3 to m^3:

V_sand_in_m3 = 508.87 dm^3 / (10 dm/m)^3
≈ 0.50887 m^3

Therefore, there are approximately 0.50887 m^3 of sand in the heap.

volume of cone is 1/3 pi r^2 h

subtract the small cone from the large cone

That's how much is sand.

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