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Question
16\sqrt{5} 45^{\circ} x = number 4 y = color © gina wilson (all things algebra), 2016
Step1: Identify the triangle type
This is a 45 - 45 - 90 right - triangle. In a 45 - 45 - 90 triangle, the legs are equal (\(x = y\)), and the hypotenuse \(c\) is related to the leg \(a\) by the formula \(c=a\sqrt{2}\). Here, \(c = 16\sqrt{5}\).
Step2: Solve for \(x\) and \(y\)
Since \(c=a\sqrt{2}\), then \(a=\frac{c}{\sqrt{2}}\). Substitute \(c = 16\sqrt{5}\) into the formula:
$$a=\frac{16\sqrt{5}}{\sqrt{2}}=\frac{16\sqrt{5}\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{16\sqrt{10}}{2}=8\sqrt{10}$$
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\(x = 8\sqrt{10}\), \(y = 8\sqrt{10}\)