McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
Mid-Chapter Quiz
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Exercise 10 Page 749

Draw a radius from the center of the circle to C. The segment BD is the side of a right triangle.

4.29

In the given diagram we can see that the length of the circle's radius is 8. In the exercise we are also given that the chord has a length of 13.5.

Circle and two perpendicular segments, one being a diameter and one a chord.
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, let's draw a radius in order to have a right triangle. Keep in mind that since the radius is constant, its length is always 8.
Circle and two perpendicular segments, one being a diameter and one a chord. A radius is added

Finally, we will pay close attention to the right triangle we have just drawn. Be aware of the fact that the chord being bisected tells us that the length of the horizontal leg is half that of the chord. Chord length/Bisected → 13.5/2= 6.75 In the right triangle the hypotenuse has a length of 8 and the horizontal leg has a length of 6.75. We want to find the length of the vertical leg.

Right triangle. The length of one leg is 6.75. The length of the hypotenuse is 8.
To find the length of BD we will substitute these values into the Pythagorean Theorem.
a^2+b^2=c^2
6.75^2+(BD)^2= 8^2
Solve for BD
45.5625+(BD)^2=64
(BD)^2=18.4375
BD=sqrt(18.4375)
BD=4.29389...
BD ≈4.29
The length of BD, rounded to the nearest hundredth, is 4.29.