Official Solution: Given isosceles triangle ABC with \(\text{base}=AC=\frac{2}{a}\) and \(\text{height}=a^2\).
\(\text{area}=\frac{1}{2}*\text{base}*\text{height}=\frac{1}{2}*\frac{2}{a}*a^2=a\). So, the question basically ask whether \(a \gt 1\)?
(1) \(\angle ABC \lt 90^\circ\). Assume \(\angle ABC=90^\circ\) then hypotenuse is \(AC=\frac{2}{a}\), as ABC becomes isosceles right triangle (\(45-45-90=1-1-\sqrt{2}\)) then the \(\text{leg}=BC=AB=\frac{2}{a \sqrt{2}}=\frac{\sqrt{2}}{a}\).
But \(BC=\frac{\sqrt{2}}{a}\) also equals to \(\sqrt{(\frac{1}{a})^2+(a^2)^2}\), so we have \(\frac{\sqrt{2}}{a}=\sqrt{(\frac{1}{a})^2+(a^2)^2}\), which leads to \(2=1+a^6\). So, finally we have that \(a^6=1\) or \(a=1\) (as \(a\) per diagram is positive). Now, if we increase \(a\) then the base \(\frac{2}{a}\) will decrease and the height \(a^2\) will increase thus making angle ABC smaller than 90 and if we decrease \(a\) then the base \(\frac{2}{a}\) will increase and the height \(a^2\) will decrease thus making the angle ABC greater than 90. So, as angle ABC is less than 90, \(a\) must be more than 1. Sufficient.
(2) Perimeter of the triangle \(ABC\) is greater than \(\frac{4}{a}\).
\(P=BC+AB+AC=2BC+AC=2\sqrt{(\frac{1}{a})^2+(a^2)^2}+\frac{2}{a} \gt \frac{4}{a}\);
\(\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a}\);
\(a^6 \gt 0\);
\(a \gt 0\). Hence we don't know whether \(a \gt 1\) is true. Not sufficient.
Answer: A
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