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(D) Let $r$ be the radius of the circle. For chord $AB$ of length $12 \, cm$,the distance from the centre $O$ is $d_1 = 8 \, cm$. $A$ perpendicular fr

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$6$

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(D) Let $r$ be the radius of the circle. For chord $AB$ of length $12 \, cm$,the distance from the centre $O$ is $d_1 = 8 \, cm$. $A$ perpendicular from the centre to a chord bisects the chord. Thus,the distance from the centre,the radius,and half the chord length form a right-angled triangle. $r^2 = (12/2)^2 + 8^2 = 6^2 + 8^2 = 36 + 64 = 100$. So,$r = 10 \, cm$. For chord $CD$ of length $16 \, cm$,let the distance from the centre be $d_2$. $r^2 = (16/2)^2 + d_2^2$. $10^2 = 8^2 + d_2^2$. $100 = 64 + d_2^2$. $d_2^2 = 36$. $d_2 = 6 \, cm$.

In a circle with centre $O$,the lengths of two chords $AB$ and $CD$ are $12 \, cm$ and $16 \, cm$ respectively. If the chord $AB$ is at a distance $8 \, cm$ from the centre,what is the distance of the chord $CD$ from the centre (in $, cm$)?

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