Exercise 17 Page 737

We want to find the length of EB in the given circle.

Since the chord and the radius are perpendicular, we know that the radius 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 14.

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 one of the legs is 11. Chord length/Bisected → 22/2=11 The hypotenuse is given to be 14. We want to find the length of the other leg AE. To find the length AE, we will substitute these values into the Pythagorean Theorem.

a^2+b^2=c^2

SubstituteValues

Substitute values

AE^2+11^2=14^2

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Solve for AE

CalcPow

Calculate power

AE^2+121=196

SubEqn

LHS-121=RHS-121

AE^2=75

SqrtEqn

sqrt(LHS)=sqrt(RHS)

AE=sqrt(75)

SplitIntoFactors

Split into factors

AE=sqrt(25* 3)

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

AE=sqrt(25)sqrt(3)

CalcRoot

Calculate root

AE=5sqrt(3)

Now we can calculate the length of EB. By the Segment Addition Postulate, the length of AB is equal to the sum of lengths AE and EB. With this information, we can write an equation that contains EB.

EB+AE=AB

SubstituteII

AE= 5sqrt(3), AB= 14

EB+ 5sqrt(3)= 14

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Solve for EB

SubEqn

LHS-5sqrt(3)=RHS-5sqrt(3)

EB=14-5sqrt(3)

UseCalc

Use a calculator

EB=5.3397459...

RoundDec

Round to 2 decimal place(s)

EB≈ 5.34