Could someone answer this for me?
what would be the lengths of the sides of an equilateral triangle formed inside a circle of diameter 16cm ?
Thanks
You want an equilateral triangle whose 3 vertices all touch the circle and whose edges are all contained within the circle.
Okay so first find the center of the circle. Now from this point, draw 3 lines. Each line should go up to and bisect a vertex of the triangle. These lines are radii of the circle. They also divide the equilateral triangle into 3 isosceles triangles.
The isosceles triangles have 2 sides made up of the radius (length 8) and the third side which is one side of the equilateral triangle. We want to solve for that side. The angles of the isosceles triangles are 120 by the center of the circle, and 30 on the other two points at the border of the circle.
Now take one of the triangles and draw a line to bisect the 120 degree angle. You now have 2 new right triangles with measurements 30-60-90. We know that sin = opp/hyp. If you take sin(60) then you'll get (half of the side of the equilateral triangle) divided by (the radius of the circle).
Thus (half of the side of the equilateral triangle) = 8 * sin(60) = 8 * sqrt(3)/2 = 4 * sqrt(3).
And then the entire side of the equilateral triangle is twice that = 8 * sqrt(3).
If you don't follow some of that then just draw what I told you to draw and you might see it more clearly.
