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?
if a right triangle is inscribed in a circle, the hypotenuse is a diameter of the circle.
so, circle radius is 6
area of circle = 36pi
area of triangle 6*6sqrt(3)
p(outside) = 1-sqrt(3)/pi = .45
To find the probability that the coin will land in the circle, but outside the triangle, we need to calculate the ratio of the areas of the regions.
First, let's find the area of the circle. The area of a circle is given by the formula: A = πr^2, where r is the radius. Since the length of the hypotenuse of the triangle is given as 12 inches, the radius of the circle will be half of that, which is 6 inches. So, the area of the circle is A = π(6^2) = 36π square inches.
Next, let's find the area of the triangle. In a 30-60-90 triangle, the ratio of the sides is 1:√3:2. Since the hypotenuse is given as 12 inches, the shorter leg will be 6 inches (√3 times smaller) and the longer leg will be 12 inches (2 times larger). The area of a triangle can be calculated using the formula: A = 0.5 * base * height. In this case, the base is 6 inches and the height is 6√3 inches. So, the area of the triangle is A = 0.5 * 6 * 6√3 = 18√3 square inches.
Now, to find the probability that the coin will land in the circle but outside the triangle, we need to find the ratio of these two areas. So, the probability can be calculated as: P = (Area of circle - Area of triangle) / Area of circle.
Plugging in the values we calculated:
P = (36π - 18√3) / 36π.
This is the probability that the coin will land in the circle, but outside the triangle.