1.) The radius which is OB also is 10.0.
 It is a right triangle so c^2 = a^2 + b^2 : In this case, AO^2 = AB^2 + OB^2
 11.4^2 = 6^2 + OB^2
 129.96 = 36 + OB^2
 OB^2 = 93.96
 OB = 9.69
 Note: I think your given is wrong since there's no 9.69 in the choices. But a similar problem is found on the internet and the given are  AB = 6 and AO = 11.7. So If these are the given data, the answer would be 10.0, Letter D.
2.) The perimeter is 64.
 In an incircle, the distances between a vertex and the two nearest tangent points are equal to each other. 
 JA = JB
 LA = LC
 KC = KB
 2 × (12 + 15 + 5) = 64
3.) The measure of ZWX is 243º.
 A diameter forms an arc of 180°, so ZRW would equal to 180°. To get ZWX, we'll add ZRW and WX.
 180° + 63° = 243°
4.) The length of AB is 58.5.
 We'll get the length of CB first, by using OB^2 = OC^2 + CB^2, substitute and transpose to get CB.
 CB = sqrt(32^2 - 13^2)
 CB = sqrt(1024 - 169) 
 CB = sqrt(855) 
 CB = 29.24
 AB is equal to 2CB.
 AB = 2(29.24) = 58.5