What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other 2 vertices on the circle?
(A)$\frac{1}{2}$
(B)1
(C)$\sqrt2$
(D)$\pi$
(E)$\frac{1+\sqrt2}{4}$
I do not know if it is a right-angled triangle or no? How can I think in this question?
Best Answer
One doesn't really need calculus for this problem. Without loss of generality, we can take the first two vertices to be $(0,0)$, $(1,0)$ and the third to be some point $(x,y)$ on the unit circle with $y>0$. The resulting triangle has base $1$ and height $y$, so that the area is $y/2$. But the largest possible value of $y$ is $1$, so the max area is $1/2$.