+52 (this is an update of the same question I already posted; I received two answers from separate users but neither worked so I'm re-posting this in hopes that someone will see this and can figure out what to do next)
Triangle XYZ is equilateral. Points Y and Z lie on a circle centered at O such that X is the circumcenter of triangle OYZ, and X lies inside triangle OYZ. If the area of the circle is 48pi, then find the area of triangle XYZ.
What I've tried:
I know that radius of the circle = OY = OZ = 4sqrt3.
Because triangle OYZ is isosceles, I drew altitude OA to meet YZ, then I know it goes through X and splits XYZ into two 30-60-90 triangles.
Following one of the hints provided in the original question, I made let 2x be the side of the equilateral triangle.
Then, AX = x(sqrt3) And because of the 30-60-90 triangle, OX = XY = 2x.
Now, I think the next thing to do is use triangle OAY or triangle OAZ to find x, probably with the Pythagorean theorem, but I'm not sure how to start that.
I've requested an extension but I still want to solve this problem ASAP. Thank you to anyone who can help!
Since X is the circumcenter of triangle OYZ, it lies on the perpendicular bisectors of all three sides of the triangle. This means that XY = XZ = R, and YZ = 2R/sqrt(3), which is the diameter of the circle.
The area of an equilateral triangle with side length s is (s^2 * sqrt(3)) / 4. So, the area of triangle XYZ is
(R^2 * sqrt(3)) / 4 = (48pi * sqrt(3)) / 4 = 12sqrt(3)pi.
Therefore, the area of triangle XYZ is 12*sqrt(3)*pi.
+36 Given that triangle XYZ is equilateral and X is the circumcenter of triangle OYZ, it implies that angle OXY is 120 degrees. This makes angle OXZ also 120 degrees, making triangle OXZ isosceles.
Since X is the circumcenter of triangle OYZ, XY = XZ. Therefore, angle XYZ = angle XZY = (180 - 120) / 2 = 30 degrees.
Now , we can use the area of the circle
A = πr2
where r is the radius of the circle centered at o.
Given that the area of the circle is 48π
we can find the radius
48π = πr2
r2 = 48
r = √48
r = 4√3
Since triangle OXY is isosceles, we can find the height of the triangle (which is also the radius of the circle):
Height = r = 4√3
Now, we can use the forumala for the area of an equilateral triangle A = (side2 * √3)/4
Area of XYZ = (XY2 * √3)/4
= (4√3)2 * √3)/4
= 48√3/4
Area of XYZ = 12√3
So, the area of triangle XYZ is 12√3 square units.
