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Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

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BuyFindarrow_forward

Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

In the figure for Exercises 21 to 24, R S ¯ is tangent to the circle at S. See Theorem 6.3.7.

Given: R S ¯ T V ¯ a n d R T = 6

Find: RS.

(HINT: Use the Quadratic Formula.)

To determine

To find:

To find RS.

Explanation

Given that, RS¯TV¯andRT=6.

The diagrammatic representation is given below,

Theorem:

If a tangent segment and a secant segment are drawn to a circle from an external point, then the square of the length of the tangent equals the product of the length of the secant with the length of its external segment.

By using the theorem 6.3.7 to get the following,

RS2=RVRT

We know that

RV=RT+TVRV=6+TVRV=6+RS. Since RS¯TV¯

Substitute the values RT=6,RV=6+RS in the above equation to get the following,

RS2=(6+RS)6RS2=36+6RSRS26RS36=0

To solve the above quadratic equation to get the following,

x=b±b24ac2a

Substitute the values of x=RS,a=1,b=6andc=36 in the above equation to get the following,

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