An equilateral triangle is inscribed in a circle of radius 4r. Express the area A within the circle but outside the triangle as a function of x, if the length of a side of the triangle is 3x
Let the equilateral triangle with side length 3x be ABC.
Since ABC is an equilateral triangle, the height (h) of the triangle is given by h = x√3.
The area of the equilateral triangle ABC is given by
Area of ABC = (1/2) * base * height = (1/2) * 3x * x√3 = (3√3 / 2)x².
Now, consider triangle OAB where O is the center of the circle and AB is one of the sides of the equilateral triangle.
Since ABC is an equilateral triangle, angle AOB is 60 degrees. Thus, triangle OAB is a 30-60-90 triangle with OA = 4r and AB = 3x.
Using the ratios of a 30-60-90 triangle, we have OB = OA / √3 = (4r) / √3.
Since BO = AB = 3x, we can express the radius of the circle in terms of x as OB = 3x = (4r) / √3.
Solving for r, we get r = (3x√3) / 4.
The area within the circle but outside the equilateral triangle ABC can be expressed as the difference between the areas of the circle and the triangle:
A = π(4r)² - (3√3 / 2)x² = π(16(3x√3)² / 16) - (3√3 / 2)x²
A = 48πx² - (3√3 / 2)x²
A = (96π - 3√3 / 2)x²
Therefore, the area within the circle but outside the equilateral triangle is given by A = (96π - 3√3 / 2)x².