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 E, B, equals, 34EB=34, 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 is a radius so EA ⟂ AB at A. In right triangle EAB, AB = sqrt(EB^2 − EA^2) = sqrt(34^2 − 16^2) = sqrt(1156 − 256) = sqrt(900) = 30.