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Explanation:
For any 45-45-90 right triangle, the hypotenuse H is always sqrt(2) times the leg L
In terms of symbols, we can say
[tex]H = L*\sqrt{2}[/tex]
We're told the hypotenuse is e, so H = e
Let's solve for L in terms of e
[tex]H = L*\sqrt{2}\\\\e = L*\sqrt{2}\\\\L*\sqrt{2} = e\\\\L = \frac{e}{\sqrt{2}}\\\\[/tex]
Now let's multiply top and bottom by [tex]\sqrt{2}[/tex] to rationalize the denominator
[tex]L = \frac{e}{\sqrt{2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{2}*\sqrt{2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{2*2}}\\\\L = \frac{e\sqrt{2}}{\sqrt{4}}\\\\L = \frac{e\sqrt{2}}{2}\\\\[/tex]
If the hypotenuse is e, then the length of each leg is [tex]\frac{e\sqrt{2}}{2}[/tex]
This matches with choice C, which is one of the two answers. This allows us to rule out choices A and B, as those expressions are not equivalent to the expression for choice C.
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The two legs of any right triangle form the base and height. The order doesn't matter. The base is always perpendicular to the height. In a right triangle, the two legs are always perpendicular.
The base and height are [tex]\frac{e\sqrt{2}}{2}[/tex] each.
The area of the triangle is...
[tex]\text{Area} = \frac{1}{2}*\text{base}*\text{height}\\\\\text{Area} = \frac{1}{2}*\frac{e\sqrt{2}}{2}*\frac{e\sqrt{2}}{2}\\\\\text{Area} = \frac{e\sqrt{2}*e\sqrt{2}}{2*2*2}\\\\\text{Area} = \frac{e*e\sqrt{2*2}}{2*2*2}\\\\\text{Area} = \frac{e^2\sqrt{4}}{2*2*2}\\\\\text{Area} = \frac{e^2*2}{2*2*2}\\\\\text{Area} = \frac{e^2}{2*2}\\\\\text{Area} = \frac{e^2}{4}\\\\[/tex]
This matches with choice E, which is the other answer. Choices D and F are eliminated as they are not equivalent to the expression for choice E.
