   Chapter 6.3, Problem 10E ### Elementary Geometry for College St...

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

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

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

# In the figure for Exercises 7 to 16, O is the center of the circle. See Theorem 6.3.5.Given: A E = 7 E B = 5 , D C = 12 Find: DE and EC. To determine

To find:

The values of DE and EC

Explanation

Given that AE=7,EB=5,DC=12

The diagrammatic representation is given below,

Theorem:

If two chords intersect within a circle, then the product of the lengths of the segments (parts) of one chord is equal to the product of the lengths of the segments of the other chord.

By using the theorem to get the following,

AEEB=DEEC...(1)

Also, know that DC=DE+EC...(2)

Substitute the values AE=7,EB=5,DC=12 in equation (1) and (2) to get the following,

AEEB=DEEC75=DEECDEEC=35DE=35EC...(3)

DE+EC=12...(4)

Substitute equation (3) in equation (4) to get the following,

DE+EC=1235EC+EC=1235+EC2=12ECEC212EC+35=0

Solve the above quadratic equation to get the following,

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