Here's my attempt to "show work" - Saab General Bulletin Board - Saabnet.com
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Picture the circle as a clock face, with an equilateral triangle inscribed inside where the corners of the triangle are at 12 o'clock, 4 o'clock, and 8 o'clock. Now draw a line from the exact center of the circle (also the exact center of the triangle) to 4 o'clock and 8 o'clock. This produces a smaller triangle with angles of 30, 30, and 120 degrees. One last line from the center towards 6 o'clock will divide this smaller triangle in half, leaving two 30-60-90 degree triangles.
It has been proven in geometry that the sides of a 30-60-90 triangle are proportional to x (opposite the 30 degree angle), 2x (opposite the 90), and [sqrt 3]x (opposite the 60). In our case, the 2x side is equal to 10 (the radius of the circle), so we can solve for x, which is 5. Therefore, the bottom edge of this small triangle is equal to 5 times [sqrt 3]. This is equal to half of one side of the equilateral triangle, so doubling this gives us 10 times [sqrt]3.
posted by 24.195.59...
Posts in this Thread:
- if you passed high school geometry read on..., Liam, Thu, 18 Nov 2004 12:56:10
- Here's another solution..., zeke, Sat, 20 Nov 2004 17:28:54
- So, the question remains....>>>, PGAero
, Thu, 18 Nov 2004 16:34:40 - Re: if you passed high school geometry read on..., Liam, Thu, 18 Nov 2004 16:11:03
- 10" x (the square root of 3), or ~17.32", IrieTom
, Thu, 18 Nov 2004 13:37:03 - Re: if you passed high school geometry read on..., Gene N
, Thu, 18 Nov 2004 13:35:29 - There's only one equilateral triangle that can be..., zeke, Thu, 18 Nov 2004 13:35:00
- Re: square that diameter and 1/2? n/m, t_mokes, Thu, 18 Nov 2004 13:11:35
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