【Explanation】: 9) In the first picture, we have a circle with points R, U, and T. RS = 24, ST = 18, and UT = x + 2. Since RS is a diameter, we can use the Pythagorean theorem to find UT. RS^2 = ST^2 + UT^2. Substituting the given values, we get 24^2 = 18^2 + (x + 2)^2. Solving this equation gives x = 10, so UT = 10 + 2 = 12.
10) TS is given as 24x, so substituting x = 10, we get TS = 240.
11) In the third picture, we have a circle with points L, M, and J. LM = 14x + 1, KL = 12, and KJ = 18. Since KL and KJ are tangent to the circle, they are perpendicular to the radius at the point of tangency. Therefore, we can use the Pythagorean theorem to find LM. LM^2 = KL^2 + JL^2. Substituting the given values, we get (14x + 1)^2 = 12^2 + 18^2. Solving this equation gives x = 3, so LM = 14*3 + 1 = 43.
12) In the second picture, we have a triangle STV with ST = 24x, TU = 27, and VU = 45. Using the Pythagorean theorem, we get ST^2 = TU^2 + VU^2. Substituting the given values, we get (24x)^2 = 27^2 + 45^2. Solving this equation gives x = 3, so ST = 24*3 = 72.
13) TV is the hypotenuse of the right triangle STV, so TV = sqrt(ST^2 + VU^2) = sqrt(72^2 + 45^2) = 87.
14) In the third picture, we have a circle with points J, K, and L. JL = 2x - 5, KL = 12, and KJ = 18. Using the Pythagorean theorem, we get JL^2 = KL^2 + KJ^2. Substituting the given values, we get (2x - 5)^2 = 12^2 + 18^2. Solving this equation gives x = 13, so JL = 2*13 - 5 = 21.
15) JK is the hypotenuse of the right triangle JKL, so JK = sqrt(JL^2 + KL^2) = sqrt(21^2 + 12^2) = 25.
16) In the fourth picture, we have a circle with points C, D, and F. CD = 45, DF = 27, and CF = 8x. Using the Pythagorean theorem, we get CF^2 = CD^2 + DF^2. Substituting the given values, we get (8x)^2 = 45^2 + 27^2. Solving this equation gives x = 9, so CF = 8*9 = 72.
【Answer】: 9) 12, 10) 240, 11) 43, 12) 72, 13) 87, 14) 21, 15) 25, 16) 72
