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
