Question

Please let me know if my answer is correct. I will attach the problem. Thank you so much!!

Answer 1

You are correct.

Answer 2

Thank you so much!!

Answer 3

Are you sure about that @mathstudent Does'nt SAS mean 2 sides and the INCLUDED angle?

Answer 4

correct me if I'm wrong

Answer 5

I see what you are talking about. I saw the two sides and over looked the fact that the side was not included in the angle! Thanks for pointing that out! So, is it SSS only, right?

Answer 6

Thank you @welshfella

Answer 7

@vic21
@welshfella

By your logic, then SSS would not work either.

The reason SSS works is that side FZ is congruent to itself, and side FZ is a side of both triangles. Since you are given the other two pairs of congruent sides, now with side FZ you have three sides congruent to three sides, and the triangles are congruent by SSS.

We know for sure you can state side FZ is congruent to itself because it is true; also, there is a theorem that states a segment is congruent to itself, and it is the only way you can have SSS. Then by the same statement of segment FZ is congruent to segment FZ, now you have SAS, not by using the two given congruent pairs of sides and the angle, that would be SSA which does not exist, but by using seg FE =~ seg FG, seg FZ =~ seg FZ, and the given included angles. That is SAS and it is correct.

The answer is still choice C, both apply, and you, @vic21, were correct in the first place.

Answer 8

yes I see what you mean though I dont see why you say SSS does not work ...

Answer 9

yes I see what you mean- the common side is one of the included angles