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The circle with center O has two chords AC and EF which are of same length 9.07.

OD and OB are the two perpendiculars drawn from the center O to the two chords AC and EF .It represents the distance of the chords from the centre.

The circle theorem states: congruent chords are equidistant from the center.

OD is congruent to OB.

Option A is the right answer.

We can conclude that [tex]\boxed{\overline {{\text{OD}}} {\text{ is congruent to }}\overline {{\text{OB}}} }[/tex]. Option (A) is correct.

Further Explanation:

Given:

The options are as follows,

A. [tex]\overline {{\text{OD}}} {\text{ is congruent to }}\overline {{\text{OB}}}.[/tex]

B. [tex]\overline {{\text{OD}}} {\text{ is congruent to }}\overline {{\text{EF}}}.[/tex]

C. [tex]\overline {{\text{OD}}} {\text{ is congruent to }}\overline {{\text{AB}}}.[/tex]

D. [tex]\overline {{\text{OD}}} {\text{ is congruent to }}\overline {{\text{OC}}}.[/tex]

Explanation:

The length of AB and EF is [tex]9.07{\text{ units}}.[/tex]

AB and EF are the chords of the circle with same length.

OB and OD are the perpendicular bisector of the chords AB and EF respectively.

The given chords of the circle are equidistant from the center of the circle.

We can conclude that [tex]\boxed{\overline {{\text{OD}}} {\text{ is congruent to }}\overline {{\text{OB}}} }[/tex]. Option (A) is correct.

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Line and Rays

Keywords: congruent, fill, blank, conclude, OH, OE, chord, tangent, displacement, two chords, radius, center, circle, equidistant,  

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