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, 11EA=11 and A, B, equals, 16AB=16, solve for E, BEB. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."

Answer 1

Because EA is a radius to the tangent point A, EA ⟂ AB. Thus triangle EAB is right-angled at A, and EB is the hypotenuse.

EB = sqrt(EA^2 + AB^2) = sqrt(11^2 + 16^2) = sqrt(121 + 256) = sqrt(377) ≈ 19.4.