RA Wrote:The relationship you referred to would hold true if the the sides were corresponding sides of similar triangles.
S1^2/S2^2 = A1/A2
in the situation referenced above, this is impossible, because both sides have the same length but "correspond to" (which, i assume, means "are placed opposite") different angles.
if you have CORRESPONDING sides of similar triangles - the sides that can form the valid proportion you wrote above - then those sides are always in
exactly the same spatial orientation in their respective triangles (i.e., opposite the same angle measures).
therefore, if you have 2 similar triangles in which CORRESPONDING sides measure 10 and 10, then the triangles have a ratio of 1:1, and are thus not only similar but congruent.
by contrast, it's possible to have two sides of the same length in
non-congruent similar triangles, but then they won't be corresponding sides (and there won't be any implications for area ratios). for instance, if i have triangles with side lengths 2, 3, 4 and 4, 6, 8, then both triangles (coincidentally) have a side of length 4; however, since it's the
proportions that matter in similar triangles, that fact is totally irrelevant.
an an analogy, model cars are "similar" (geometrically speaking) to the real cars. having congruent sides (like the 4's above) is sort of like this: let's say an entire model car is 10.2 inches long, and say the accelerator pedal of the real car is 10.2 inches long. this is a nice coincidence, but it's of no importance whatsoever.
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on to the original post.
Can we compare sides and angles of two different triangles ? Say for example if I know that in one traingle a smaller angle corresponds to side equal to 10 and in another triangle a greater angle corresponds to 10 , can I safely say that the the one with the smaller angle would have a greater area ?
no.
example:
look at
the following picture, showing a right triangle inscribed in a semicircle.
i can put the vertex of that right angle ANYWHERE on the semicircular arc, and it will still be a 90° angle, and it will still be across from the exact same diameter. but the areas will change - from near zero (if the vertex is located close to one of the endpoints of the diameter) to a maximal value (if the vertex is located directly at the top, at the midpoint of the semicircle).
sorry about that; it was a nice idea :(