McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
7. Special Segments in a Circle
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Exercise 15 Page 772

Notice that the 9 inches long segment is the perpendicular to the 12 inches chord. Also, use the Segments of Chords Theorem.

13 inches

Practice makes perfect

Let's make a diagram to illustrate the given information.

Notice that CD passes through the center of the circle, which means it lies on a diameter. Also, because CD⊥AB, we get that CD bisects AB. Then, AC=CB=6 inches.

Next, if E is the other endpoint of the diameter DE, we can apply the Segments of Chords Theorem to set the following equation. DC* CE = AC* CB Let's substitute the corresponding measures to find CE.
DC* CE = AC* CB
9* CE = 6* 6
â–Ľ
Solve for CE
9* CE = 36
CE = 4
Finally, we calculate the diameter of the circle by using the Segment Addition Postulate. d = DC + CE ⇒ d &= 9+4 ⇒ d &= 13 In consequence, the diameter of the original cake was 13 inches.