odot(O, 41) and odot(O, 9) are concentric cir | vedclass

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(D) Let $O$ be the common center of the two concentric circles. The radius of the larger circle is $OB = 41$ and the radius of the smaller circle is $

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) Let $O$ be the common center of the two concentric circles. The radius of the larger circle is $OB = 41$ and the radius of the smaller circle is $OM = 9$.Since the chord $\overline{AB}$ of the larger circle is tangent to the smaller circle at $M$,the radius $OM$ is perpendicular to the chord $AB$ at the point of tangency $M$.Thus,in the right-angled triangle $\Delta OMB$,by the Pythagorean theorem:$OB^2 = OM^2 + MB^2$$41^2 = 9^2 + MB^2$$1681 = 81 + MB^2$$MB^2 = 1681 - 81 = 1600$$MB = \sqrt{1600} = 40$.Since the perpendicular from the center to a chord bisects the chord,$AB = 2 	imes MB = 2 	imes 40 = 80$.

$\odot(O, 41)$ and $\odot(O, 9)$ are concentric circles. The chord $\overline{AB}$ of $\odot(O, 41)$ touches $\odot(O, 9)$ at point $M$. Then $AB = \ldots$

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