08 Mar '07 13:49

Can a 24 inch-long flute fit into a box measuring 12 inches by 14 inches by 16 inches? If no, why not. If yes, explain.

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Grammar dyslexic08 Mar '07 14:07

Not without taking it apart or snapping it in half. I think the longest line inside the box from bottom corner to its opposite top corner would be 21.26 inches. I'm probably missing something, but I'm going with my frist instinct ðŸ™‚*Originally posted by wittywonka***Can a 24 inch-long flute fit into a box measuring 12 inches by 14 inches by 16 inches? If no, why not. If yes, explain.**- Joined
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09 Mar '07 16:19

The answer is yes. And it is nicely shown.*Originally posted by wittywonka***Can a 24 inch-long flute fit into a box measuring 12 inches by 14 inches by 16 inches? If no, why not. If yes, explain.**

Part 2 of the same problem is:

What diameter can the flute have in order to fit in that box?

(Suppose that the flute is cylindrical.)

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10 Mar '07 00:28

That's not hard.*Originally posted by wittywonka***Alrighty, here's yet another geometry problem, much harder. What is the circumference of a circle inscribed in an equilateral triangle with a height of (12 * square root of 3)?**

The inscribed circle of an equilateral triangle is centred on the triangle's centre. The triangle's centre is situated 1/3rd of the way along each of the triangle's three height lines. The radius of the circle is equal to the distance of the centre, along the height line, to the nearest side of the triangle. All these are easily seen when you draw a diagram, and flow elementarily from the equilaterality of the triangle.

The radius of the circle is, then, a third of 12*sqrt(3), is 4*sqrt(3). It circumference is therefore 2 pi r, equals 2 pi 4 sqrt(3), equals 8 pi sqrt(3). Its area is more interesting; that's pi r**2, which is pi 4**2 sqrt(3)**2, equals pi * 16 * 3, equals 48 pi, eliminating the square root.

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10 Mar '07 14:58

i'll be glad if you show me the answer and how to do it*Originally posted by FabianFnas***The answer is yes. And it is nicely shown.**

Part 2 of the same problem is:

What diameter can the flute have in order to fit in that box?

(Suppose that the flute is cylindrical.)

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10 Mar '07 15:021 editnevermind i think i know how to do it

at the rectangle box the corner is where the flute ends

the flute is allowed with 24.4131112315...and it's 24in long. therefore the allowed space on each side is .4131112315/2 in long.

when the diameter cuts, the leftover area become a rect prism with side lenth RATIO of 12:14:16 cuted in half. therefore .4131112315 is the diagnal of rectangle diamention ratio of 6:7:8.

since the maxium diameter creates a circle that slices the little rect prism in half, it's diameter is equal to the diagnal of the little cube. so the diameter maxium allowed is 0.4131112315 inches, about 1 centimeter.