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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 6.3, Problem 23E
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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
1 views

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.)

Chapter 6.3, Problem 23E, In the figure for Exercises 21 to 24,RS is tangent to the circle at S. See Theorem 6.3.7. Given:

To determine

To find:

To find RS.

Explanation

Given that, RS¯≅TV¯ and RT=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=RVâ‹…RT

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+6RSRS2−6RS−36=0

To solve the above quadratic equation to get the following,

x=−b±b2−4ac2a

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

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