Answer: [tex]8\sqrt{2}[/tex]
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Work Shown:
Focus entirely on triangle ABD (or on triangle BCD; both are identical)
The two legs of this triangle are AB = 8 and AD = 8. The hypotenuse is unknown, so we'll say BD = x.
Apply the pythagorean theorem.
[tex]a^2 + b^2 = c^2\\\\c = \sqrt{a^2 + b^2}\\\\x = \sqrt{8^2 + 8^2}\\\\x = \sqrt{2*8^2}\\\\x = \sqrt{8^2*2}\\\\x = \sqrt{8^2}*\sqrt{2}\\\\x = 8\sqrt{2}\\\\[/tex]
So that's why the diagonal BD is exactly [tex]8\sqrt{2}\\\\[/tex] units long
Side note: [tex]8\sqrt{2} \approx 11.3137[/tex]
