In right △ABC , the right angle is at C, m∠A=30∘ , and AC=5√ 5 units.
What is the perimeter of △ABC ?
Using the 30-60-90 right triangle theorum,
the hypothenuse is 5 sqrt(5)
BC is the 5 sqrt(5) / 2
AC is then 5 sqrt(15) / 2
add them up and you get
15 sqrt(5) / 2 + 5 sqrt(15) / 2
