A piece of paper is randomly selected from a stack that contains 15 small, 44 large, and 20 extra large pieces.
1 answer:
You might be interested in
Answer:
Exact height = 8*sqrt(3) mm
Approximate height = 13.856 mm
=============================================================
Explanation:
If you do a vertical cross section of the cylinder, then the 3D shape will flatten into a rectangle as shown in the diagram below.
After flattening the picture, I've added the points A through F
- point A is the center of the sphere and cylinder
- points B through E are the corner points where the cylinder touches the sphere
- point F is at the same horizontal level as point A, and it's on the edge of the cylinder.
Those point labels will help solve the problem. We're told that the radius of the sphere is 8 mm. So that means segment AD = 8 mm.
Also, we know that FA = 4 mm because this is the radius of the cylinder.
Focus on triangle AFD. We need to find the height x (aka segment FD) of this triangle so we can then double it later to find the height of the cylinder. This in turn will determine the height of the bead.
------------------------------
As the hint suggests, we'll use the pythagorean theorem
a^2 + b^2 = c^2
b = sqrt(c^2 - a^2)
x = sqrt(8^2 - 4^2)
x = sqrt(48)
x = sqrt(16*3)
x = sqrt(16)*sqrt(3)
x = 4*sqrt(3)
This is the distance from D to F
The distance from D to E is twice that value, so DE = 2*(FD) = 2*4*sqrt(3) = 8*sqrt(3) is the exact height of the bead (since it's the exact height of the cylinder).
Side note: 8*sqrt(3) = 13.856 approximately.
