Question

in the given fi gure, O is centre of the circle. If OP = 13 cm, QP = 24 cm and OR is perpendicular to QP, then

fi nd the length of SR. ​

Answer 1

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GiveN :

$\mapsto\sf OP=13\: cm$

$\mapsto\sf QP=24\: cm$

$\mapsto\sf OR \:\:is\:\: perpendicular \:\:to \:\:QP$

SolutioN :

$\sf As \:\:OR\:\:is\:\: perpendicular\:\:to \:\:QP$

$\therefore \sf \triangle OPS \:\: is \:\:a \:\:right \:triangle \:\:and \:\:S\:\:is \:\:the \:\:mid-pt \:\:of\:\:QP$

$\sf Hence,$

$\sf{{PS} }=\frac{{{PQ}}}{2}$

$\implies \sf {{PS}}=12$

Now,

$\sf PS^2+SO^2=PO^2\:\:\:\:\:\:\bf [\:Pythagoras\:\: Theorem\:]$

$\implies \sf 12^2+OS^2=13^2$

$\sf\implies OS=5$

Now,

$\sf OP=13\:\:\bf [Given]$

hence,

$\sf OR=13\:\:\:\bf [∵OR=OP]$

$\sf\implies OS+SR=13$

$\sf \implies 5+SR=13$

$\:\:\:\:\:\:\therefore {\underline{\boxed{\sf SR=8}}}$

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HOPE THIS IS HELPFUL...