Consider a circle with center O and radius OA=9 cm. Let AB be a chord of length 12 cm.
We see that OA and OB are two radii of the circle: OA=OB=9 cm.
We have the equilateral triangle ΔAOB:OA=OB.
AD is the height of the triangle AOB. Then AD is the perpendicular bisector and
AD=DB=21AB Consider the right triangle OAD
The Pythagorean Theorem
OA2=OD2+AD2OD=OA2−AD2
OA=9 cm,AD=21(12 cm)=6 cm
OD=(9 cm)2−(6 cm)2=35 cm The perpendicular distance from the center of the circle to the chord is 35 cm.
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