Little explorations with HP calculators (no Prime)
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04-07-2017, 05:41 PM
(This post was last modified: 04-07-2017 05:48 PM by Gerson W. Barbosa.)
Post: #138
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RE: Little explorations with the HP calculators
(04-07-2017 05:19 PM)Han Wrote:(04-07-2017 12:57 PM)SlideRule Wrote: IF the base and the height remain constant, the AREA of the triangle is also constant, irrespective of an alteration of the "shape" of the original triangle Yes, we can! In fact that was my first approach. I've found cos  = -1/4, cos B = 7/8 and cos Č = 11/16, the sides of the inner triangle being 2sqrt(5/8)x, sqrt(3/2)x and sqrt(47/8)x. Incidentally the area of the inner triangle = 7. AB and AC must not be collinear. That's the only restriction I am aware of. |
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