Question

Given circle E with diameter CD and radius EA. AB is tangent to E at A. If AC = 9 and AD = 4, solve for CD. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."​

Answer 1

Since AB is tangent to circle E at A, we know that angle CAB is a right angle (tangent is perpendicular to the radius at the point of tangency). Therefore, triangle ADC is a right triangle.

Let's use the Pythagorean theorem to find the length of CE:

CE^2 = AC^2 + AE^2 (using Pythagorean theorem in triangle ACE)

CE^2 = 9^2 + EA^2 (since AE = EA, by definition of radius)

CE^2 = 81 + EA^2

We still need to find EA. Let's use the fact that EA is half the length of CD:

EA = CD/2

Now we can substitute this expression into the previous equation:

CE^2 = 81 + (CD/2)^2

CE^2 = 81 + CD^2/4

Next, let's use the Pythagorean theorem in triangle ADC:

AD^2 + DC^2 = AC^2

4^2 + DC^2 = 9^2

DC^2 = 9^2 - 4^2

DC^2 = 65

Now we can substitute this expression into the previous equation:

CE^2 = 81 + 65/4

CE^2 = 99.25

Taking the square root of both sides, we get:

CE ≈ 9.96

Therefore, CD = 2CE ≈ 19.9.

Answer: CD ≈ 19.9