Could someone walk me through/show me their work for this one? I think I messed up around applying the 30-60-90 rules. I said that CH would equal 3root8 over 2 because I had AC equal to 3root8.
Start by noticing:
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AD = 16 because ABCD is a parallelogram.
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\angle ACD = 90^{\circ} because \overline{AB} \parallel \overline{\rm CD}.
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\angle ADC = 60^{\circ} because opposite angles in a parallelogram are congruent.
Can you work out CD from \triangle ACD?
Finally, you can repeat this process on \triangle CHD, where \angle D = 60^{\circ}, \angle HCD = 30^{\circ}, and \angle CHD = 90^{\circ}.