Find area of triangle in half circle !enter image description here Half circle with $r = 2$. So, think $CD = BD = 2, CEB$ = right triangle Area = $ 1/2 \overline{CE}\ \overline{AB}$ What is the max area? By theory max area is when $h = r$ but how to prove it?

Hint:

$$Area = \frac{1}{2}bh$$

where $b$ is the base of the triangle and $h$ is the height. As $r=2$, the diameter must be $4$ and so the base of the triangle must be $4$. To find the maximum height, observe that the triangle would be the largest when $h=r$, as you suggested. So, consider

$$Area = \frac{1}{2}bh = \frac{1}{2}(2r)(r)=\frac{1}{2}(4)(2)=4$$

or observe that the maximum area is

$$r^2 = 2^2 =4$$