Question

The length of the chord of the larger circle which is tangent to the smaller circle of two concentric circles is 'a' units. Then the area enclosed between the concentric circle is​

Answer 1

Answer:

hey here is your solution

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refer the reference diagram uploaded above

Step-by-step explanation:

$To \: find = \\ area \: enclosed \: between \: two \: circles \: (in \: terms \: of \: a) \\ \\ so \: here \: AB \: is \: a \: chord \: of \: larger \: circle \\ so \: AB = a.units \\ \: \\ so \: here \: let \: the \: common \: centre \: of \: two \: concentric \: circles \: be \: O \\ so \: OM \: is \: a \: perpendicular \: drawn \: from \: centre \: O \: on \: chord \: of \: larger \: circle \: AB \\ so \: we \: know \: that \\ perpendicular \: drawn \: from \: centre \: of \: circle \: on \: its \: chord \: bisects \: the \: chord \\ \\ hence \: then \\ AM=MB = \frac{1}{2} \: OM \: \: \: \: \: \: \: \: \: (1)\\ \\ now \: AM+MB = AB \: \: \: \: \: (A - M - B \: ) \\ AM+AM = AB \: \: \: \: \: \: \: \: \: \: from \: (1) \\ \\ ie \: AB = 2.MB \\ ie \: MB = \frac{AB}{2} \\ \\ = \frac{a}{2} \: units \: \: \: \: \: \: \: (2)$

$hence \: then \\ from \: (1) \: and \: (2) \\ we \: get \\ MB = \frac{OM}{2} \\ \\ = \frac{a}{2} = \frac{OM}{2} \\ \\ hence \: as \: well \: OM = a.units \: \: \: \: \: \: \: (3)$

$now \: conidering \: △ \: OMB \\ as \: \: ∠ OMB = 90 \\ by \: applying \: pythagoras \: theorem \\ we \: get \\ OM {}^{2} \: + \: MB {}^{2} = OB {}^{2} \\ = (a) {}^{2} + ( \frac{a}{2} ) {}^{2} \\ \\ = a {}^{2} \: + \: \frac{a {}^{2} }{4} \\ \\ = \frac{4a {}^{2} + a {}^{2} }{4} \\ \\ hence \: \: OB {}^{2} = \frac{5a {}^{2} }{4}$

$now \: thus \: then \\ OB \: is \: a \: radius \: of \: larger \: circle \: and \\ OM \: is \: that \: of \: smaller \: circle \\ \\ so \: here \: OM {}^{2} = a {}^{2} \\ OB {}^{2} = \frac{5a {}^{2} }{4} \\ \\ so \: area \: of \: smaller \: circle = \pi.OM {}^{2} \\ = \pi.a {}^{2} \: \: \: \: \: \: \: \: (4) \\ \\ similarly \: area \: of \: larger \: circle = \pi.OB {}^{2} \\ \\ = \frac{\pi.5a {}^{2} }{4} \: \: \: \: \: \: \: \: (5)$

$so \: thus \: then \\ area \: enclosed \: between \: two \: circles = \\ area \: of \: large \: circle \: - \: area \: of \: smaller \: circle \\ \\ = \frac{\pi.5a {}^{2} }{4} \: - \: \pi.a {}^{2} \\ \\ = \pi( \frac{5a { }^{2} }{4} \: - \: a {}^{2} \: ) \\ \\ = \pi( \frac{5a {}^{2} - 4a {}^{2} }{4} \: ) \\ \\ = \pi( \frac{a {}^{2} }{4} \: ) \\ \\ = \frac{\pi.a {}^{2} }{4} \: units {}^{2}$