Problem: If â–³ABC is inscribed in a semicircle whose diameter is AB, then AC+BC must be:
(Hint: a⩾b is read " a is equal to or greater than b.")
Answer Choices:
A. equal to AB
B. equal to AB2​
C. ⩾AB2​
D. ⩽AB2​
E. AB2
Solution:
The inscribed triangle with maximum perimeter is the isosceles right triangle; in this case AC+BC=AB2​. Therefore, in general, AC+BC≤AB2​.