Four balls 4 inches in diameter are placed in a square box whose inside base dimensions are 8 inches. In the space between the first four balls a fifth of the same diameter is placed. How deep must the box be in order that the top will just touch the fifth ball?
2006-08-17 22:07:33 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Nagpapatawa ba kayo?
The box must be 4 + 2 sqrt(2) inches deep.
2006-08-17 22:12:06 · answer #1 · answered by Joe Mkt 3 · 2⤊ 0⤋
Draw a pyramid with the centers of all five spheres as vertices.
Each edge of this pyramid is equal to 4
The height of a triangular face of this pyramid is sqrt(12)
The height of the pyramid is sqrt(8) = 2*sqrt(2)
The total height of all five spheres is half the diameter of one sphere plus the height of the pyramid plus half the diameter of one sphere:
2 + 2sqrt(2) + 2 = 4 + 2sqrt(2) inches
2006-08-18 05:11:16 · answer #2 · answered by Pablo Fanques 3 · 1⤊ 0⤋
6 inches
2006-08-18 05:13:42 · answer #3 · answered by Ättitude is Ëverything 3 · 0⤊ 1⤋
6 inches
2006-08-18 05:13:10 · answer #4 · answered by Indian 2 · 0⤊ 1⤋
It is 4 + 2(squareroot(2))
2006-08-18 05:31:44 · answer #5 · answered by heidavey 5 · 1⤊ 0⤋
6.something
2006-08-18 05:51:43 · answer #6 · answered by bsc504 3 · 0⤊ 1⤋
12-2sqrt2
2006-08-18 05:29:48 · answer #7 · answered by Anonymous · 0⤊ 1⤋
good question !!!
wild a** guess says 6.78 inches
2006-08-18 05:23:05 · answer #8 · answered by atheistforthebirthofjesus 6 · 1⤊ 0⤋
whaaaaaaaaaaatttttttt????!!!!!!!!
2006-08-18 06:38:37 · answer #9 · answered by ?????????? 2 · 0⤊ 1⤋