17 < \(\alpha\) < 18
To solve this problem, let's first understand the properties of a right-angled isosceles triangle. In this triangle, the two legs are equal, and the hypotenuse is equal to the leg length multiplied by the square root of 2. Let's denote the length of each equal side (leg) as \(a\).
Given that the perimeter of the triangle is 20 units, we can write the equation:
\(2a + a\sqrt{2} = 20\)
We need to solve this equation to find \(a\). Rearrange the terms:
\(2a + a\sqrt{2} = 20 \Rightarrow a(2 + \sqrt{2}) = 20\)
Solve for \(a\):
\(a = \frac{20}{2 + \sqrt{2}}\)
To simplify the expression, multiply the numerator and the denominator by the conjugate of the denominator:
\(a = \frac{20(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})}\)
\((2 + \sqrt{2})(2 - \sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2\)
Thus,
\(a = \frac{20(2 - \sqrt{2})}{2} = 10(2 - \sqrt{2})\)
Now, we calculate the area \(\alpha\) of the triangle, which is given by:
\(\alpha = \frac{1}{2} \times a \times a = \frac{1}{2} \times a^2\)
Substitute the value of \(a\):
\(a^2 = (10(2 - \sqrt{2}))^2 = 100(4 - 4\sqrt{2} + 2) = 100(6 - 4\sqrt{2}) = 600 - 400\sqrt{2}\)
So, the area is:
\(\alpha = \frac{1}{2} \times (600 - 400\sqrt{2}) = 300 - 200\sqrt{2}\)
Calculate an approximation:
\(\sqrt{2} \approx 1.414\)
\(300 - 200 \times 1.414 = 300 - 282.8 = 17.2\)
The area \(\alpha\) is approximately 17.2, which falls in the option \(17 < \alpha < 18\).
Therefore, the correct answer is \(17 < \alpha < 18\).
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