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Little explorations with HP calculators (no Prime)
04-07-2017, 02:52 PM (This post was last modified: 04-07-2017 02:53 PM by Gerson W. Barbosa.)
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RE: Little explorations with the HP calculators
(04-07-2017 12:57 PM)SlideRule Wrote:  
(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote:  ...The shape of the outer triangle is irrelevant ...

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

Exactly! As long as the base and the height are kept constant, the area won't change.

(04-07-2017 12:57 PM)SlideRule Wrote:  
(04-07-2017 02:29 AM)Gerson W. Barbosa Wrote:  ... the sides of the outer triangle = 2x, 3x and 4x...
Then why not give the original dimensions entirely in terms of x? The substitution is plausable; is it implied in the original set of givens?

Although apparently misleading, that might have excluded the right-triangle possibility, which is perhaps the most adequate way to solve the problem (at least using trigonometry). An interesting variation of the problem might be the case CF = BF. The solution is also an integer number.

Gerson.
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RE: Little explorations with the HP calculators - Gerson W. Barbosa - 04-07-2017 02:52 PM



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