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, 13EA=13 and D, B, equals, 7DB=7, 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 B as on the secant through C and D so we can use the tangent–secant (power of a point) theorem:

EA = 13 ⇒ CD = 2·EA = 26.
BD = 7 ⇒ BC = BD + DC = 7 + 26 = 33.

Tangent–secant: (BA)^2 = BD·BC = 7·33 = 231, so
BA = √231 ≈ 15.2.

Answer: AB ≈ 15.2.