Answer to Question #126885 in Geometry for jungkookiee

Question #126885
A chord of length 12 cm is drawn in a circle of radius 9 cm. Calculate the perpendicular distance from the center of the circle to the chord.
1
Expert's answer
2020-07-19T15:26:26-0400

Consider a circle with center OO and radius OA=9 cm.OA=9\ cm. Let ABAB be a chord of length 12 cm.12\ cm.



We see that OAOA and OBOB are two radii of the circle: OA=OB=9 cm.OA=OB=9 \ cm.

We have the equilateral triangle ΔAOB:OA=OB.\Delta AOB: OA=OB.

ADAD is the height of the triangle AOB.AOB. Then ADAD is the perpendicular bisector and


AD=DB=12ABAD=DB={1\over 2}AB

Consider the right triangle OADOAD

The Pythagorean Theorem


OA2=OD2+AD2OA^2=OD^2+AD^2OD=OA2AD2OD=\sqrt{OA^2-AD^2}

OA=9 cm,AD=12(12 cm)=6 cmOA=9 \ cm, AD=\dfrac{1}{2}(12\ cm)=6\ cm


OD=(9 cm)2(6 cm)2=35 cmOD=\sqrt{(9\ cm)^2-(6\ cm)^2}=3\sqrt{5}\ cm

The perpendicular distance from the center of the circle to the chord is 35 cm.3\sqrt{5}\ cm.


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Comments

Assignment Expert
19.07.20, 22:26

A solution of the question has already been published.

jungkookiee
19.07.20, 11:31

Why not answered?

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