Summer Problem Solving Marathon Solution #13

If we drop perpendiculars from A and B to side DC, hitting that side at E and F respectively, then AEFB is a rectangle of length 5 and unknown width.

Triangle BFC is then a 45-45-90 right triangle. The sides of a 45-45-90 right triangle are in the ratio 1-1-√2. In this case, the hypotenuse BC is 3√2, so the two legs are each of length 3. Thus BF is 3, and so AF and FB must also be 3.

Triangle AED is a 30-60-90 right triangle. The sides of a 30-60-90 right triangle are in the ratio 1-√3-2. In this case, the side AE, which is opposite the 60 degree angle, is of length 3. So side DE, opposite the 30 degree angle, is of length 3/√3, which simplifies to √3.

Side DC is then the sum of DE, EF, and FC. That sum is 5 + 3 + √3 = 8 + √3.