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, 8EA=8 and E, B, equals, 17EB=17, solve for A, BAB. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."
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, 8EA=8 and E, B, equals, 17EB=17, solve for A, BAB. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."
Solution
The problem involves a circle and a tangent line, which forms a right triangle. We can use the Pythagorean theorem to solve for AB.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This can be written as: a² + b² = c²
In this case, EB is the hypotenuse, EA is one side, and AB is the other side. We are trying to solve for AB.
Given: EA = 8 (radius of the circle) EB = 17 (hypotenuse)
We can plug these values into the Pythagorean theorem:
(8)² + AB² = (17)² 64 + AB² = 289 AB² = 289 - 64 AB² = 225
Taking the square root of both sides to solve for AB gives:
AB = √225 AB = 15
So, the length of AB is 15 units.
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