The area of the square ABCD is 16cm². The bisector BT of triangle ABD intersects the diagonal

The area of the square ABCD is 16cm². The bisector BT of triangle ABD intersects the diagonal AC at point O. Calculate the length of the radius of the circle around triangle AOD.

Find the side of the square and its diagonal:
a = √S = 4 cm.
BD = √ (2 * a²) = 4 * √2 = 5.65685.

Consider an isosceles triangle BOD:
<OBD = <ODB = 45 ° / 2 = 22.5 °.
<BOD = 180 ° – 45 ° = 135 °.

By the sine theorem, we find the lengths of OD and AO:
OD = BD * sin (22.5 °) / sin (135 °) = 3.0615;
AO = OD * sin (22.5 °) / sin (45 °) = 1.6568.

We calculate the semiperimeter of the triangle AOD:
p = 0.5 * (4 + 3.0615 + 1.6568) = 4.3592.

Find the radius of the circumscribed circle:
R = (4 * 3.0615 * 1.6568) / (4 * √ (4.3592 * (4.3592 – 4) * (4.3592 – 3.0615) * (4.3592 – 1.6568) )) = 2.1648.

Answer: the radius of the circumscribed circle is 2.16 cm.