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 19 Page 772

Start with evaluating the value of a using Theorem 10.17.

a=15, b≈ 11.3

Practice makes perfect

Consider the given diagram.

We will find the values of a and b one at a time.

Finding a

Let's focus on the tangent and the secant.

Let's remember Theorem 10.17, which tells us about the segments of secants and tangents.

Theorem 10.17

If a tangent and a secant intersect in the exterior of a circle, then the square of the measure of the tangent is equal to the product of the measures of the secant and its external secant segment.

According to the theorem, the square of the measure of the tangent will be equal to the product of the measures of the secant and its external secant segment.
4( 4+a+6)= 10^2
Solve for a
4(10+a)=10^2
40+4a=10^2
40+4a=100
4a=60
a=15
The value of a is 15.

Finding b

Now let's find the value of b.

We will start by recalling Theorem 10.15, which tells us about two intersecting chords.

Theorem 10.15

If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

In our diagram, two chords intersect in the interior of a circle. Therefore, the products of the lengths of the chord segments are equal.
15* 6 = b* 8
Solve for b
90=8b
90/8=b
45/4=b
11.25=b
b=11.25
b≈ 11.3
The value of b is approximately 11.3.