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MathRevolution
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==> In the original condition, for a right triangle, there are 2 variables (2 legs) 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), you get 6:8:10 and the perimeter of the right triangle becomes 6+8+10=24, hence unique and sufficient.

Therefore, the answer is C.
Answer: C
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Question : Can you please elaborate more the answer? How did you get the 6 and 8?
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MathRevolution
What is the perimeter of a certain right triangle?

1) The hypotenuse’s length is 10
2) The triangle’s area is 24

The trick with this question is that in statement 1 we cannot actually assume that the sides are 6 and 8 just because we know that

x^2 +y^2 = 100

We cannot make any assumptions about X and Y UNLESS there is a restriction such as "X and Y must be integers" or "The product of XY is 48."

C
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Question : Can you please elaborate more the answer? How did you get the 6 and 8?

Assume we have two sides \(a\) and \(b\) and a hypothesis 10.
Then we have \(a^2 + b^2 = 100\) and \(\frac{1}{2}ab = 24\)
Thus \(a = 6\), \(b = 8\) or \(a = 8\), \(b = 6\).
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