Question

radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord.​

Answer 1

Answer:

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Answer 2

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt Radius\: of \:a \:circle \:is\: 34 \:cm\: and\: the\: distance\\\tt of\: the \:chord \:from\: the\: centre\: is\: 30 \:cm, \:find \:the\:\\\tt length\: of \:the \:chord.$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\sf \bf O \:is\:the \:centre \:of \:the \:circle$
• $\sf \bf Radius \:of\:the\:circle\Big(OA \:and \:OB\Big) =34 \:cm$
• $\sf \bf Distance \:of\:chord \:from \:the \:centre\Big(OC\Big) =30 \:cm$
• $\sf \bf OC \perp AB$

$\sf \bf {\boxed {\mathbb {TO\:FIND}}}$

• $\sf \bf Length \:of \:the \:chord$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

$\sf \bf In \triangle OCA$

$\sf \bf \angle C={90}^{\circ}$

${\underbrace {\overbrace {\color {orange} {\tt By\: pythagoras \:theorem}}}}$

$\sf \bf \implies {OA}^{2}={AC}^{2}+{CO}^{2}$

$\sf \bf \implies {\Big(34\Big)}^{2}={AC}^{2}+{\Big(30\Big)}^{2}$

$\sf \bf \implies 1156={AC}^{2}+900$

$\sf \bf \implies {AC}^{2}=1156-900$

$\sf \bf \implies {AC}^{2}=256$

$\sf \bf \implies AC=\sqrt{256}$

$\implies {\boxed {\color{green} {\sf \bf AC=16\:cm}}}$

$\sf \bf In \triangle OCB$

$\sf \bf \angle C={90}^{\circ}$

${\underbrace {\overbrace {\color {orange} {\tt By\: pythagoras \:theorem}}}}$

$\sf \bf \implies {OB}^{2}={BC}^{2}+{CO}^{2}$

$\sf \bf \implies {\Big(34\Big)}^{2}={AC}^{2}+{\Big(30\Big)}^{2}$

$\sf \bf \implies 1156={BC}^{2}+900$

$\sf \bf \implies {BC}^{2}=1156-900$

$\sf \bf \implies {BC}^{2}=256$

$\sf \bf \implies BC=\sqrt{256}$

$\implies {\boxed {\color{green} {\sf \bf BC=16\:cm}}}$

------------------------------------------------------------

$\sf \bf \implies AB=AC+BC$

$\sf \bf \implies AB=16+16$

$\implies {\boxed{\boxed {\color{blue} {\sf \bf AB=32\:cm}}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \bf\huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\sf \bf Circumference \:of \:the \:circle =2\pi{r}^{2}$

$\sf \bf Area \:of \:the \:circle =\pi{r}^{2}$

$\sf \bf Diameter\:of \:the \:circle =2r$