If my attached image is correct, then based on my computation NT is equal to C. 12.05
S is the center of the circle. OM = 18.
Based on the attached image, I concluded the following facts:
1) RM is the diameter
2) OM = NM ; NM = 18
3) ST and SU are radii so ST = SU
4) Q is the midpoint of SU ; QU = 8
5) P is the midpoint of ST. PT = QU thus PT = 8
6) P is the midpoint of NM thus NP = 9
Points NPT creates a right triangle. NT is the hypotenuse. We will use the Pythagorean theorem to solve for NT.
a² + b² = c²
9² + 8² = c²
81 + 64 = c²
145 = c²
√145 = √c²
12.0415 = c
S is the center of the circle. OM = 18.
Based on the attached image, I concluded the following facts:
1) RM is the diameter
2) OM = NM ; NM = 18
3) ST and SU are radii so ST = SU
4) Q is the midpoint of SU ; QU = 8
5) P is the midpoint of ST. PT = QU thus PT = 8
6) P is the midpoint of NM thus NP = 9
Points NPT creates a right triangle. NT is the hypotenuse. We will use the Pythagorean theorem to solve for NT.
a² + b² = c²
9² + 8² = c²
81 + 64 = c²
145 = c²
√145 = √c²
12.0415 = c