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Question

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. The length of common chord   is


A
6 cm
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B
4 cm
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C
12 cm
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D
8 cm
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Solution

The correct option is B 6 cm
$$ Given-\\ A\quad \& \quad B\quad are\quad two\quad intersecting\quad circles,\quad the\quad comon\quad chord\quad of\quad which\quad is\quad \\ CD\quad .\\ CD\quad intersects\quad AB\quad or\quad extended\quad AB\quad at\quad E.\\ AB=4cm,\quad radius\quad of\quad the\quad circle\quad with\quad centre\quad A\quad is\quad AC=5cm\\ and\quad radius\quad of\quad the\quad circle\quad with\quad centre\quad B\quad is\quad BC=3cm.\\ To\quad find\quad out-\\ The\quad length\quad of\quad the\quad common\quad chord\quad CD.\\ Solution-\\ \angle AEC={ 90 }^{ o }\quad since\quad the\quad line\quad joining\quad the\quad centres\quad of\quad of\quad two\quad \\ intersecting\quad circles\quad is\quad perpendicular\quad to\quad their\quad common\quad chord.\\ So\quad we\quad have\quad two\quad right\quad triangles,\quad \Delta ACE\quad \& \quad \Delta BCE\quad with\quad \\ hypotenuses\quad AC\quad \& \quad BC\quad respectively.\quad \\ Let\quad BE=xcm.\\ Then\quad AE=(4+x)cm\quad and\quad BE=xcm.\\ \therefore \quad By\quad Pythagoras\quad theorem,\quad we\quad get\\ CE=\sqrt { AE^{ 2 }-{ AC }^{ 2 } } =\sqrt { { BC }^{ 2 }-{ BE }^{ 2 } } \\ \Longrightarrow \sqrt { { 5 }^{ 2 }-{ \left( 4+x \right)  }^{ 2 } } =\sqrt { { 3 }^{ 2 }-{ x }^{ 2 } } \\ \Longrightarrow x=0.\\ i.e\quad B\quad \& \quad E\quad are\quad same\quad point.\\ \therefore \quad AB=AE=4cm.\\ So\quad CE=\sqrt { AE^{ 2 }-{ AC }^{ 2 } } =\sqrt { { 5 }^{ 2 }-{ \left( 4+0 \right)  }^{ 2 } } cm=3cm.\\ Now\quad CD=2CE\quad since\quad the\quad line\quad joining\quad the\quad centres\quad of\quad of\quad two\quad intersecting\quad circles\\ bisects\quad the\quad common\quad chord.\\ \therefore \quad The\quad length\quad of\quad the\quad common\quad chord\quad CD=2\times 3cm=6cm.\\ Ans-\quad Option\quad A.\\  $$
286320_242482_ans.png

Mathematics

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