Let's first determine the radius (r) of the circle. Since the triangle is isosceles with a 60-degree angle at the center, it is an equilateral triangle, and each side is 12 meters long. Now, let's split the circle into two parts and use the Pythagorean theorem.
The equilateral triangle can be divided into two 30-60-90 triangles. In a 30-60-90 triangle, the side opposite the 30-degree angle is half the length of the hypotenuse. Therefore, half of the side length of the triangle (12m) is equal to 6m.
Now using the Pythagorean theorem, we can find the radius (r) of the circle:
r^2 = hypotenuse^2 - (6m)^2
r^2 = (12m)^2 - (6m)^2
r^2 = 144m^2 - 36m^2
r = sqrt(108m^2) = 6 sqrt(3)m
The area of the circle is:
Area_circle = pi * r^2 = pi * (6 sqrt(3)m)^2 = 108m^2 * pi
Now, let's find the area of the equilateral triangle:
Area_triangle = (side^2 * sqrt(3)) / 4
Area_triangle = (12^2 * sqrt(3)) / 4
Area_triangle = (144 * sqrt(3)) / 4
Area_triangle = 36 sqrt(3)m^2
Now we can find the area of the shaded region:
Area_shaded = Area_circle - Area_triangle = (108 * pi * m^2) - (36 sqrt(3)m^2)
So the area of the shaded region is (108 * pi - 36 sqrt(3)m^2). This matches answer choice (c).
Answer: (c) (120*pi + 36*sqrt(3))m^2
What is the area of the shaded region in the given circle in terms of pi and in the simplest form?
the picture shown is of a circle with a triangle from 1 oclock to 3 oclock with a 60 deg opening. the side length of the triangle is 12m and the little part where the triangle and the edge of the circle meet between 1 and 3 oclock is the only part not shaded. The choices we have as answers are...
a) (120pie+6sqrt3)m^2
b) (96pie+36sqrt3)m^2
c)(120pie+36sqrt3)m^2
d) (96pie+6sqrt3)m^2
1 answer