Question

Given circle EE with diameter start overline, C, D, end overline
CD
and radius start overline, E, A, end overline
EA
. start overline, A, B, end overline
AB
is tangent to EE at AA. If E, A, equals, 16EA=16 and D, B, equals, 18DB=18, solve for A, BAB. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."

Answer 1

AB = 30.

Reason: EA = 16 so diameter CD = 2·16 = 32. For external point B, power of a point gives BA^2 = BD·BC. The secant through B meets the circle at D and C, so BC = BD + DC = 18 + 32 = 50. Hence BA^2 = 18·50 = 900, so BA = sqrt(900) = 30.