   Chapter 6.4, Problem 32E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

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

Circle O has a diameter of length 20 c m . Chord A B ¯ has length 12 c m , and chord C D ¯ has length 10 c m . How much closer is A B ¯ than C D ¯ to point O ?

To determine

To find:

The distance that AB¯ is closer than CD¯ to point O.

Explanation

Given:

Circle O has a diameter of length 20cm. Chord AB¯ has length 12cm, and chord CD¯ has length 10cm.

Theorem Used:

According to the Pythagorean theorem,

hypotenuse2=base2+perpendicular2.

Property Used:

The perpendicular bisector divides the chord into two equal parts.

Approach Used:

i) Use the Pythagorean theorem to calculate the distance of chords from center i.e. PM¯ and PN¯.

ii) Find the difference between PM¯ and PN¯.

Calculation:

The chords equals AB¯=12, CD¯=10 and the radii equals OB¯=OC¯=202=10

In the given figure,

Let M and N be perpendicular bisector on chords AB¯ and CD¯ respectively.

Applying Pythagorean theorem,

hypotenuse2=base2+perpendicular2

In AOB,

OB¯2=OM¯2+MB¯2 or OM¯=OB¯2MB¯2

And from COD,

OC¯2=ON¯2+NC¯2

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