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Big Ideas Math Geometry, 2014

BI

Big Ideas Math Geometry, 2014 View details

6. Segment Relationships in Circles

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Exercise 7 Page 572

Use the Segments of Secants and Tangents Theorem. Then solve a quadratic equation by factoring, completing the square, or some other method.

x=8

Practice makes perfect

We are given the following diagram and asked to find the value of x.

To do this we will use the Segments of Secants and Tangents Theorem. Let's recall what it states.

Segments of Secants and Tangents Theorem
If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the lengths of the secant segment and its external segment equals the square of the length of the tangent segment.

First, we need to calculate the lengths of the secant SP by adding x and 10. SP=x+10 According to this theorem, the product of SP and TP is equal to the square of PR. SP* TP =PR^2 Let's substitute SP with x+10, TP with x, and PR with 12 and solve the equation for x.

SP* TP =PR^2

SubstituteValues

Substitute values

( x+10)( x) = 12^2

Calculate power

(x+10)(x)=144

Distribute x

x^2+10x=144

This is a quadratic equation, so let's solve it by completing the square. Note that all terms with x are on one side of the equation, so we do not need to rewrite it. In this equation, b=10. Using this information, we can calculate ( b2 )^2.

( b/2 )^2

b= 10

( 10/2 )^2

Calculate quotient

5^2

Calculate power

25

Next, we will add ( b2 )^2=25 to both sides of our equation. Then, we will factor the trinomial on the left-hand side, and solve the equation.

x^2+10x=144

▼

Complete the square and Solve for x

LHS+25=RHS+25

x^2+10x+ 25=144+ 25

FacPosPerfectSquare

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

(x+5)^2=144+25

Add terms

(x+5)^2=169

sqrt(LHS)=sqrt(RHS)

sqrt((x+5)^2)=sqrt(169)

Calculate root

x+5=± 13

LHS-5=RHS-5

x=- 5± 13

We conclude that the equation has two solutions. x_1&=- 5+13=8 x_2&=- 5-13=- 18 On the diagram, x represents the distance TP, so it cannot have a negative value. Therefore, x is equal to 8.

Subchapter links

Communicate Your Answer p.569

Monitoring Progress p.570-572