All the given points are collinear, so we have ... AB +BC = AC BC +CS +SD = BD The problem statement tells us BD = AC BC +CS +SD = AB +BC . . . . . substituting for BD and AC BC +CS +2 = 4 +BC . . . . . . . . substituting given lengths CS = 2 . . . . . . . . . . . . . . . . . . . subtract (2+BC)
CS is the radius of Circle C, so its diameter is twice that: 4. The radius of Circle D is the diameter of Circle C plus SD, so is 4 +2 = 6. The diameter of Circle D is twice that: 12.
The diameter of Circle C is 4. The diameter of Circle D is 12.
Step-by-step explanation: you should switch both into the same denominator for the fractions so that would turn into 3/12 and 8/12 then add them up to get 11/12 so he has 1/12 left