A 30-60-90 triangle is inscribed in a circle. The length of the hypotenuse is 12 inches. If a coin is tossed on the figure, what is the probability that the coin will land in the circle, but outside the triangle?

Answer 1

All right-triangles inscribe in a circle with a diameter equal to the hypotenuse.

Therefore for the 30-60-90 triangle, the radius of the circle is 6 inches, and the short side is also 6 inches. The height is 6sqrt(3), so the area of the triangle is
At=36sqrt(3)/2 = 18 sqrt(3) sq.in.

The area of the circle is
Ac=π6^2=36π
The probability of falling inside the circle and outside the triangle is therefore
P(C\T)=(Ac-At)/Ac

Answer 2

the quesiton is land in the circle, but outside the triangle. Is it the probability is (AC - AT)/AC
81.22/113 x 100%
= 72%

Answer 3

Correct.
I get
P(C\T) (probability inside circle minus triangle)
=(Ac-At)/Ac
=81.92/113.1=72.4%