Final answer:
To solve for the diameter CD of circle E, use the Pythagorean theorem on the right triangle formed by the radius EA, the tangent segment AC, and the hypotenuse CE (the radius extension to the circle's circumference) and then double the result because the diameter is twice the radius.
Step-by-step explanation:
The question involves geometry and specifically relates to the properties of circles and tangents. Given a circle E with diameter CD and radius EA, and that AB is tangent to circle E at point A, with AC being 28 and EA (which is the radius) being 16, we are asked to solve for the diameter CD.
Since EA is the radius of the circle and AB is tangent to the circle at A, the triangle EAC is a right triangle by the tangent-radius theorem.
According to the Pythagorean theorem, the square of the length of the hypotenuse (which is CE in this case) is equal to the sum of the squares of the lengths of the other two sides. Thus,
CE^2 = EA^2 + AC^2.
Plugging in the provided lengths, we have CE^2 = 16^2 + 28^2. Calculating this gives CE^2 = 256 + 784 = 1040. Taking the square root of both sides gives us CE = √1040, which is approximately 32.249031.
Since CD is the diameter and CE is the radius, CD is twice the length of CE. Therefore, CD = 2 × CE = 2 × 32.249031 ≈ 64.498062 cm. Hence, the diameter CD is approximately 64.5 cm.