Heres the whole story problem.
A sector is sliced from a donut that has a radius of 6 cm and a thickness of 2.5 cm. The central angle of the sector is 60 degrees. The curved section of the sector is sliced off, leaving the largest possible triangluar wedge. What is the volume of the wedge?
2007-01-22 13:28:00 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Yes it is the largest triangle wedge created without curves in the "sector" yeah the terms through me off as well, but i didnt write the story problem.
2007-01-22 14:10:49 · update #1
First off, donuts are round; hence, pi x r2.
Calculate volume of entire donut first. This means volume of cylinder minus the volume of the hole (which is 6 -2.5, or 3.5 cm.) You need to calculate BOTH volumes to do this, ha ha (yeesh!)
60 degrees is 1/6 of the circle, if that is what they mean by "sector." On the other hand, the "largest possible triangular wedge" would a 179 degree thing.
This much I know, but the terms are throwing me off.
2007-01-22 13:43:52 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Do you mean the largest bona-fide frustum (no curves) that can be cut from this donut sector, with end sides at 60 degrees? Good problem, can be solved, but too much work to do on a wild goose chase if this isn't what you mean.
One more question: Are all the 4 small edges of the frustum supposed to be aligned with the center of the torus, or they don't have to? Resutls are different depending on this.
2007-01-22 21:58:34 · answer #2 · answered by Scythian1950 7 · 0⤊ 0⤋
simple ,wedge multiply 2.5 x 6 x ? sorry i cant figure out the last number to multiply just try to fid the last number to multiply these two.
2007-01-22 21:38:17 · answer #3 · answered by Anonymous · 0⤊ 0⤋