In a circle with radius 17 , cm,the length of | vedclass

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(A) Let the circle have centre $O$ and radius $r = 17 \, cm$. Let $AB$ be the chord of length $30 \, cm$. Draw a perpendicular $OM$ from the centre $O

Vedclass · Accepted Answer

$8$

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(A) Let the circle have centre $O$ and radius $r = 17 \, cm$. Let $AB$ be the chord of length $30 \, cm$. Draw a perpendicular $OM$ from the centre $O$ to the chord $AB$. The perpendicular from the centre of a circle to a chord bisects the chord. Therefore,$AM = MB = \frac{AB}{2} = \frac{30}{2} = 15 \, cm$. In the right-angled triangle $	riangle OMA$,by the Pythagorean theorem: $OA^2 = OM^2 + AM^2$. Substituting the values: $17^2 = OM^2 + 15^2$. $289 = OM^2 + 225$. $OM^2 = 289 - 225 = 64$. $OM = \sqrt{64} = 8 \, cm$. Thus,the distance of the chord from the centre is $8 \, cm$.

In a circle with radius $17 \, cm$,the length of a chord is $30 \, cm$. Find the distance of the chord from the centre. (in $, cm$)

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