The two red lines are at a right angle, you can connect them and solve for the hypotenuse using the normal a^2 + b^2.
At this point you now have a second triangle that contains the X you want. Also, since the outer shape is a quarter-circle (assumed) you know that the corner in the bottom right is 90 degrees which makes the two side equal in length.
Since it’s an equilateral triangle, and you know the hypotenuse, you can back it out with c^2 = 2a^2 and solve it that way.
The two red lines are at a right angle, you can connect them and solve for the hypotenuse using the normal a^2 + b^2.
At this point you now have a second triangle that contains the X you want. Also, since the outer shape is a quarter-circle (assumed) you know that the corner in the bottom right is 90 degrees which makes the two side equal in length.
Since it’s an equilateral triangle, and you know the hypotenuse, you can back it out with c^2 = 2a^2 and solve it that way.