Answer: [tex]x = \frac{7\sqrt{6}}{2}\\\\[/tex]
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Explanation:
Focus on the upper triangle. This is a 30-60-90 triangle. The short leg is 7 and the long leg is [tex]7\sqrt{3}[/tex]. The long leg is always sqrt(3) times that of the short leg (and this only works for 30-60-90 triangles).
In short, the horizontal length is [tex]7\sqrt{3}[/tex] units long.
Now move to the lower triangle. This is a 45-45-90 triangle with congruent legs (each are x units long). We just found the hypotenuse is [tex]7\sqrt{3}[/tex]
We can use the pythagorean theorem to say the following:
[tex]a^2 + b^2 = c^2\\\\x^2 + x^2 = (7*\sqrt{3})^2\\\\2x^2 = 147\\\\x^2 = \frac{147}{2}\\\\x = \sqrt{\frac{147}{2}}\\\\x = \frac{\sqrt{147}}{\sqrt{2}}\\\\x = \frac{\sqrt{147}*\sqrt{2}}{\sqrt{2}*\sqrt{2}}\\\\x = \frac{\sqrt{147}*\sqrt{2}}{2}\\\\x = \frac{\sqrt{147*2}}{2}\\\\x = \frac{\sqrt{294}}{2}\\\\x = \frac{\sqrt{49*6}}{2}\\\\x = \frac{\sqrt{49}*\sqrt{6}}{2}\\\\x = \frac{7\sqrt{6}}{2}\\\\[/tex]
