Three equal right-isosceles triangles are inscribed in a circle. They are positioned around an equilateral triangle which stabilises the construction. The triangles are cut from A4 paper, with dimensions noted. The right triangles have two sides of length 8.3″ = 210 mm, and the equilateral triangle has side length 7.44″ = 188 mm. The six outermost points are equidistant from the centre of the equilateral triangle, thus lying on a circle, which is not shown.
Assume that the bottommost right triangle has a leg that is horizontal (as you’ve drawn it. Then, for both of the vertices of that leg to be on the circle, the center of the circle must be directly above the midpoint of that leg; i.e., in Cartesian coördinates, they must have the same x-coördinate. The line through the center of the equilateral triangle (the center of the circle) and the upper right node on that right triangle makes a 15° angle with the horizontal. A bit of trigonometry and geometry (which I can describe in detail if you like) gives us this relationship between the length of the side of the equilateral triangle (Se) and the length of the leg of the right triangle (Sr):
Se = Sr × (√3) / (2 × cos(15°)) ≈ Sr × 0.896575.
In your example, Sr = 210, so Se ≈ 210 × 0.896575 ≈ 188.28
Well done.