In the following figure,if AB = 15,then CD = | vedclass

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(B) Let the two concentric circles have radii $R$ and $r$ with center $O$. Let $OM$ be the perpendicular from $O$ to the chord $AB$ of the outer circl

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$15$

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(B) Let the two concentric circles have radii $R$ and $r$ with center $O$. Let $OM$ be the perpendicular from $O$ to the chord $AB$ of the outer circle. Since $AB$ is tangent to the inner circle at $M$,$OM = r$. In the right-angled triangle $	riangle OMA$,by Pythagoras theorem,$OA^2 = OM^2 + AM^2$,so $R^2 = r^2 + AM^2$. Since the perpendicular from the center to a chord bisects the chord,$AM = AB / 2 = 15 / 2 = 7.5$. Thus,$R^2 - r^2 = AM^2 = (7.5)^2$. Similarly,for the chord $CD$ of the outer circle,let $ON$ be the perpendicular from $O$ to $CD$. Since $CD$ is tangent to the inner circle at $N$,$ON = r$. In the right-angled triangle $	riangle ONC$,$OC^2 = ON^2 + CN^2$,so $R^2 = r^2 + CN^2$. Thus,$CN^2 = R^2 - r^2 = (7.5)^2$. Therefore,$CN = 7.5$. Since the perpendicular from the center bisects the chord,$CD = 2 	imes CN = 2 	imes 7.5 = 15$.

In the following figure,if $AB = 15$,then $CD = \ldots$

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