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==> In the original condition, for the right triangle, there are 2 variables and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2) and assume the two hypotenuses of the triangle as a and b, you get \(a^2+b^2=4^2=16\), and from ab/2=4.5, you get ab=9. Then, from \((a+b)^2=a^2+b^2+2ab=16+2(9)=16+18=34\), you get a+b=√34, so the perimeter of the right triangle=a+b+4=√34+4, hence it is unique and sufficient.

Therefore, the answer is C.
Answer: C
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