Question

the radius of a circle is 6cm and the length of one of its chords is 6cm find the distance of the chord from the centre​

Answer 1

Answer:

$\sf{The \ distance \ between \ chord \ and \ center}$

$\sf{is \ 3\sqrt3 \ cm}$

Given:

$\sf{\leadsto{Radius \ of \ circle \ is \ 6 \ cm}}$

$\sf{\leadsto{Length \ of \ one \ of \ it's \ chord \ is \ 6 \ cm}}$

To find:

$\sf{The \ distance \ of \ the \ chord \ from \ the}$

$\sf{center.}$

Solution:

$\sf{Let \ O \ be \ center \ and \ AB \ be \ the}$

$\sf{required \ chord \ and \ OP \ represents}$

$\sf{the \ distance \ between \ center \ and \ chord.}$

$\sf{Note:}$

$\sf{A \ perpendicular \ drawn \ from \ the \ centre}$

$\sf{of \ the \ circle \ bisects \ the \ chord.}$

$\sf{Here, In \ \triangle OPA}$

$\sf{OA=6 \ cm... It's \ radius}$

$\sf{AP=3 \ cm...it's \ half \ the \ length \ of \ chord.}$

$\sf{\angle OPA=90^\circ...It's \ perpendicular \ to}$

$\sf{chord \ AB.}$

$\sf{By \ Pythagoras \ theorem}$

$\sf{OA^{2}=OP^{2}+AP^{2}}$

$\sf{\therefore{6^{2}=OP^{2}+3^{2}}}$

$\sf{\therefore{36=OP^{2}+9}}$

$\sf{\therefore{OP^{2}=36-9}}$

$\sf{\therefore{OP^{2}=27}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{OP=3\sqrt3 \ cm}$

$\sf\purple{\tt{\therefore{The \ distance \ between \ chord \ and \ center}}}$

$\sf\purple{\tt{is \ 3\sqrt3 \ cm}}$