ykpgal wrote:
S = (BC∙AC)/2
BC = 1/2∙AB,
=>
AB² = BC² + AC² => AB² = 1/4∙AB² + AC²
AC = (√3)/2∙AB
=>
S = ((1/2∙AB)∙(√3/2∙AB))/2
S = (√3/8)AB²
Um, no idea what you did here. I see S = the area, but it looks like you didn't find a value for the area.
I also don't see how some people have mentioned that the base of the triangle lies along the diameter. All we know is that the triangle is inscribed inside the circle.
I got C. I don't know how to find the area, but I know it's possible.
St 1 tells us it's a right triangle, since the square of one side, equals the sum of the squares of the other two sides. But as there could be many different sizes of right traingles that would fit insiade this circle, this statement is insufficient.
St2 gives us another one of the angles. Like statement one, alone it is insufficent, but together we know know this is a 30-60-90 triangle. This will fit into this circle only one way. I don't care that I can't figure out the area, only that I know it's possible.