Given circle E with diameter CD and radius E A. AB is tangent to E at A.
If EA = 8 and EB = 18, solve for AB. Round your answer to the nearest
tenth if necessary. If the answer cannot be determined, click "Cannot be determined."

The tangent to a circle is perpendicular to the radius at the point of tangency. Given that AB is tangent to circle E at point A, the angle between the radius EA and the tangent AB is 90°. Consequently, triangle AEB forms a right triangle with EB as the hypotenuse, and EA and AB as its legs.

To find the length of leg AB, we can use the Pythagorean Theorem which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs. Therefore:

[tex]EB^2=EA^2+AB^2[/tex]

Given that EA = 8 and EB = 18:

[tex]18^2=8^2+AB^2\\\\324=64+AB^2\\\\AB^2=260\\\\AB=\sqrt{260}\\\\AB=16.124515496...\\\\AB=16.1\; \sf (nearest\;tenth)[/tex]

So, the length of segment AB is:

[tex]\Large\boxed{\boxed{AB=16.1}}[/tex]

Given circle E with diameter CD and radius E A. AB is tangent to E at A. If EA = 8 and EB = 18, solve for AB. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."

Answer :

Other Questions

Given circle E with diameter CD and radius E A. AB is tangent to E at A.
If EA = 8 and EB = 18, solve for AB. Round your answer to the nearest
tenth if necessary. If the answer cannot be determined, click "Cannot be determined."