Question 1191597
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Use the Pythagorean Theorem with legs d and hypotenuse {{{sqrt(12+d)}}}:<br>
{{{d^2+d^2=(sqrt(12+d))^2}}}
{{{2d^2=12+d}}}
{{{2d^2-d-12=0}}}<br>
That quadratic does not factor in integers, leading to an ugly answer to the question.  Since the rest of the problem is just laborious computation, I leave it to you to finish.<br>
Or perhaps check the given information to see if you have shown it incorrectly, so that the answer turns out to be "nice"....<br>