Little explorations with HP calculators (no Prime)

04072017, 02:29 AM
(This post was last modified: 04072017 05:15 AM by Gerson W. Barbosa.)
Post: #125




RE: Little explorations with the HP calculators
Area = 7
....... Spoiler ........ ........ The shape of the outer triangle is irrelevant. At first I had made the sides of the outer triangle = 2x, 3x and 4x. Then I used the law of cosines on each side of outer triangles (sides proportional to 3, 4 and 5) in order to determine the angles. By applying the law of the cosines again on these angles and the sides of the inner triangle opposed to them, these sides could be expressed in terms of x. The Heron's formula was then used to compute the areas of both triangles. Finally, the ratio between these two areas is used to find the answer to the problem. But, since the shape of the triangle is irrelevant, why not choosing a righttriangle (now I can do it :), one with angles 90, 60 and 30 degrees and sides 2x, 4x and 2x*sqrt(3)? This makes the solution even more simple: there's no need to use the Heron's formula to calculate the area of the outer triangle; also, the sides of the inner triangle are more easily obtained. Done by hand, except the evaluation and simplification of the area of the inner triangle using the Heron's formula (HP 50g CAS). 

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