Find the third side of a Triangle

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The discussion centers on finding the length of the third side of an isosceles triangle given two sides of lengths 3 cm and 7 cm. Participants clarify that the third side must be between 4 cm and 10 cm, leading to the conclusion that it can only be 7 cm, as 3 cm is below the acceptable range. The triangular inequality is highlighted as the simplest method for this determination, while some suggest using the cosine law for a more complex approach. There is a consensus that visualizing the triangle configurations aids in understanding which side lengths are possible. The conversation concludes with appreciation for the insights shared.
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Where did I go wrong?
 
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From where do you know that that angle is 60 degrees? We have no explicit information on the angles of the triangle, so how did you infer it from the given data?

Instead try to think what are the possible values for the third side (hint: the triangle is isosceles) and try to find a reason to reject one value so you ll be left with the other.
 
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Delta2 said:
(hint: the triangle is isosceles)
So of the four potential answers, only two of them should be considered.
 
The other side can be between, (7+3=)10 and (7-3=)4 cm, being an isosceles triangle, the third side should be either 7 or 3. As 3 is below the range so 7cm is the length of the other side. Thanks for the pointer.
 
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TytoAlba95 said:
The other side can be between, (7+3=)10 and (7-3=)4 cm, being an isosceles triangle, the third side should be either 7 or 3. As 3 is below the range so 7cm is the length of the other side. Thanks for the pointer.
Very well i find your reasoning correct. You used the triangular inequality. I had another reasoning in my mind using the cosine law. Let me know if you want to hear it.
 
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Delta2 said:
I had another reasoning in my mind using the cosine law.
Which is fine, but requires calculation that probably can't be done in one's head. A variation of the technique used by the OP is to draw two isosceles triangles: one with two sides of 3 units, and the other with two sides of 7 units. Pretty clearly the one with a pair of sides of 3 units can't also have a side of 7 units, but the one with a pair of 7 unit sides can have a third side of 3 units.
 
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Mark44 said:
Which is fine, but requires calculation that probably can't be done in one's head. A variation of the technique used by the OP is to draw two isosceles triangles: one with two sides of 3 units, and the other with two sides of 7 units. Pretty clearly the one with a pair of sides of 3 units can't also have a side of 7 units, but the one with a pair of 7 unit sides can have a third side of 3 units.
It is not that hard. Take a triangle isosceles with two sides of 3. We can prove using the cosine law that the third side has an upper bound of 6.
 
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Delta2 said:
It is not that hard. Take a triangle isosceles with two sides of 3. We can prove using the cosine law that the third side has an upper bound of 6.

That's not exactly the simplest way to prove 6 is the upper bound.
 
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Office_Shredder said:
That's not exactly the simplest way to prove 6 is the upper bound.
No it isnt. Using the triangular inequality is probably the simplest way.
 
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I actually needed the simplest, or more precisely the shortest method to answer this question, as I was preparing for a MCQ-based exam.
Thank you everyone for your pointers and inputs. Thank you Delta2. :)
 
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