Exercise 124 Page 721

a Recall the Segments of Chords Theorem.

Segments of Chords Theorem
If two chords intersect in a circle, then the products of the lengths of the chord segments are equal.

Using this theorem we can write an equation containing the given lengths in the diagram.

Let's solve this equation for x.

6x=7(3)

Multiply

6x=21

DivEqn

.LHS /6.=.RHS /6.

x=3.5

b Note that AD is a diameter of the circle. Since AB is 5 units then AD must be double this, which is 10 units. With this information we can determine the length of DF.

DF+5+x=AD

Substitute

AD= 10

DF+5+x= 10

SubEqn

LHS-5=RHS-5

DF+x=5

SubEqn

LHS-x=RHS-x

DF=5-x

Let's add the length of DF to the diagram.

Recall the Perpendicular Chord Bisector Theorem.

Perpendicular Chord Bisector Theorem
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.

This means AD bisects EC. Therefore, it must be that EF≅ FC. Let's add this to the diagram.

Now we can use the Segments of Chords Theorem to write an equation containing x. (5+x)(5-x)=4(4) Let's solve this equation for x.

(5+x)(5-x)=4(4)

ExpandDiffSquares

(a+b)(a-b)=a^2-b^2

25-x^2=4(4)

▼

Solve for x

Multiply

25-x^2=16

SubEqn

LHS-25=RHS-25

- x^2=-9

MultEqn

LHS * (-1)=RHS* (-1)

x^2=9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x=±3

x > 0

x=3

c We have been given expressions for MN and PQ. We also know that MN≅ PQ. Let's add this information to the diagram.

Since MN and PQ are congruent, we can set their expressions equal to each other and solve for x.

10x-8=7x+13

▼

Solve for x

SubEqn

LHS-7x=RHS-7x

3x-8=13

AddEqn

LHS+8=RHS+8

3x=21

DivEqn

.LHS /3.=.RHS /3.

x=7

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Exercise 124 Page 721

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