Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
2. Chords and Arcs
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Exercise 15 Page 777

Extend the segment that goes from the center of the circle to the chord, and draw a diameter. If the chord and the diameter are perpendicular, then the diameter bisects the chord.

20.8

Practice makes perfect

In the given diagram, we can extend the segment that goes from the center of the circle to the chord, and draw a diameter.

Since the chord and the diameter are perpendicular, we know that the diameter bisects the chord. This fact will help us find lengths later in the exercise.

Now, we will pay close attention to the right triangle formed by the chord, the diameter and the radius. Be aware of the fact that the chord being bisected tells us that the length of one of the legs is x2. Chord length/Bisected → x/2 The other leg's length is 6 and the hypotenuse is 12. We want to find the value of x.

To find the value of x, we will substitute these values into the Pythagorean Theorem.
a^2+b^2=c^2
(x/2)^2 + 6^2= 12^2
Solve for x
x^2/4 + 36 = 144
x^2/4 = 108
x^2 = 432
x = sqrt(432)
x = 20.78460...
x≈ 20.8