Question

AB and CD are two parallel chords of a circle , Which arc an oppoite Sides of the centre , Such that AB = 10 cm, CD =24 cm and the distance between AB and CD is 17 cm . Find the radius of the Circle .​

Answer 1

Answer:

answer is :- 13cm.

please see the attached picture ❤

Answer 2

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$\large\bf \pink{Answer}$

$\small \bf \red{Given = } { AB \: and \: CD \: are \: two \: parellel \: Chords \: of \: a \: circle \: , Which \: arc \: on \: opposite \: Side \: Of \: the \: centre.}$

$\small \bf{AB = \: 10 cm \: , CD \: = 24 cm}$

$\small \bf{Distance \: Between \: AB \: and \: CD \: = 17 cm}$

$\bf \red{To \: Find} \small \bf{ \: = Radius ?}$

$\small \bf\red{Construction}= \small{Draw \: OP \perp AB \: and \: OQ \perp CD \: . \: Join \: OB \: and \: OD}$

$\small\bf \red{Procedure =} \small{ Since AB \: || \: CD and OP \perp AB , OQ \perp \: CD}$

$\small \bf{ \therefore \: Point \: P, \: O \: and \: Q \: are \: Collinear.}$

$\small \bf{ Let \: \: \: \: \: OP \: = x \: cm}$

$\small \bf{Then , OP = x cm } \\\small \bf \: {OQ = ( 17- x ) cm}$

$\small\bf {PB = \frac{10}{2} = 5 \: cm }$

$\small\bf{ \because \: \perp From \: the \: centre \: bisects \: the \: chord }$

$\small \bf{QD = \frac{24}{2} = 12 cm}$

$\small \bf{In rt. \triangle OPB} \: \: \: \: \: \: \: r^{2} = x^{2} + 5^{2} ....(1)$

$\small \bf{In \: rt. \: OQD , \: \: r^{2} = (17 - x)^{2} + 12^{2} }$

$\small \bf{From (1) and (2) , We \: have}$

$\small\bf {x^{2} + 25 = (17 - x)^{2} + 12^{2} }$

$\small \bf{ \implies \: x^{2} +25 = 289 + x^{2} - 34x + 144 }$

$\small \bf{ \implies \: x^{2} - x^{2} + 38x = 289 + 144 - 25 }$

$\small \bf{ \implies \: 34x = 408}$

$\small \bf{ \implies \: x = 12 \: cm}$

$\small \bf{Using \: the \: Value \: of \: x \: in \: (1) \: we get}$

$\small \bf{r^{2} = 12^{2} + 5^{2} = 144 + 25 = 169}$

$\small \bf \red{ \therefore \: Radius \: (r) = 13 \: cm}$

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