Question 1014187
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AB is a chord of a circle, centre O, with radius 17cm.If AB=16cm, find the distance of O from AB.
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Draw the perpendicular from the center O to the cord AB, and let the point C represents its intersection with the chord AB.

The length |OC| is the distance of interest, d.

Notice that the point C bisects the chord AB. It is because the triangle OAB is isosceles having congruent sizes OA and OB (radii).

Now from the right-angled triangle OAC you have 

d = {{{sqrt(abs(OC)^2 - ((abs(AC))/2)^2)}}} = {{{sqrt(17^2 - 8^2)}}} = {{{sqrt(289 - 64)}}} = {{{sqrt(225)}}} = 15.

Thus the distance d is 15 cm.

<U>Answer</U>. The distance of O from AB is 15 cm.
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