Question

because all 45°-45°-90° triangles are _______, we can set up a proportion as well to solve for c.
\\(\frac{1}{\sqrt{2}} = \frac{8}{\\___}\\)
\\(1 \cdot c = 8\sqrt{2}\\)
\\(c = \sqrt{}
all three answers that we now have are identical, regardless of the method we chose to solve.

Explanation:

Step1: Identify Triangle Similarity

All \(45^\circ\text{-}45^\circ\text{-}90^\circ\) triangles are similar (by AA similarity, as all have two \(45^\circ\) angles and a right angle). So the first blank is "similar".

Step2: Solve the Proportion

Given \(\frac{1}{\sqrt{2}}=\frac{8}{c}\) (let the denominator be \(c\), the hypotenuse). Cross - multiply: \(1\cdot c = 8\cdot\sqrt{2}\), so the second blank (the denominator) is \(c\), and from \(c = 8\sqrt{2}\), the third blanks: the first is \(8\) and the second is \(2\).

Answer:

First blank: similar; Second blank (denominator): \(c\) (or \(8\sqrt{2}\) when solved); Third blanks: \(8\), \(2\) (for \(c = 8\sqrt{2}\))

(Note: For the proportion \(\frac{1}{\sqrt{2}}=\frac{8}{c}\), solving gives \(c = 8\sqrt{2}\), so the denominator in the proportion is \(8\sqrt{2}\), and \(c = 8\sqrt{2}\) means the first blank in \(c=\underline{8}\sqrt{\underline{2}}\) is \(8\) and the second is \(2\).)