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 A, B, equals, 25AB=25 and E, B, equals, 31EB=31, solve for E, AEA. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."

Answer 1

Because AB is tangent at A, EA (the radius) is perpendicular to AB, so triangle EAB is right at A. By the Pythagorean theorem:

EA^2 + AB^2 = EB^2

EA = sqrt(EB^2 − AB^2) = sqrt(31^2 − 25^2) = sqrt(961 − 625) = sqrt(336) = 4√21 ≈ 18.3.

So EA ≈ 18.3.