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, 18EA=18 and D, B, equals, 12DB=12, 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

Interpret the problem as asking for the length AB. Let E be the center, so EA = 18 implies the diameter CD = 2·18 = 36. If B lies on the extension of diameter CD beyond D, then BC = BD + DC = 12 + 36 = 48.

By the power-of-a-point (tangent-secant) theorem, (BA)^2 = BD·BC = 12·48 = 576, so BA = sqrt(576) = 24.

Answer: AB = 24 (or 24.0).