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, 7EA=7 and D, B, equals, 4DB=4, solve for A, BAB. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."
A
D
C
E
B
Answer
Attempt 1 out of 2

Answer 1

AB is a tangent from B, and the secant through B meets the circle at D and C. By the power of a point,
AB^2 = BD · BC = BD · (BD + DC).
Here BD = 4 and DC = diameter = 2·EA = 14, so
AB^2 = 4 · (4 + 14) = 4 · 18 = 72,
AB = sqrt(72) = 6√2 ≈ 8.5.

Answer: 8.5