A cone has a base radius of 3 and a height of \(8\) A cube is inscribed in the cone, so that four of its vertices lie in the circular base, and the other four vertices touch the lateral surface. Find the side length of the cube.
I hope this is correct.
SInce the height is 8, and the diameter is 6 (which is greater than 4), I think the cube has to end halfway up the height. This means that the answer is 4. I'm not very confident about my answer, so someone else can try to double-check it.
I think we can solve this with similar triangles
Call the side length of the cube, s
The diagonal across the cube's base = s * sqrt (2)
1/2 of this = s / sqrt 2
We can construct a triangle with a base = radius of cone - s/sqrt 2 = 3 - s/sqrt 2
And the height of this triangle = s
And the other triangle has a base of the radius of the cone = 3 and a height of 8
And these triangles wil be similar....so.....
3/ 8 = (3 - s/sqrt 2) / s cross-multiply
3s = 8 (3 - s /sqrt 2)
3s + 8s / sqrt 2 = 24
s ( 3 + 8/sqrt 2) = 24
s = 24 / (3 + 8/sqrt 2) ≈ 2.77