Question

please answer me fast in notebook and I will mark you brainliest

don't write waste answer​

Answer 1

$\\ \tt \:A \: chord \: CD \: is \: drawn \: which \: is \: parallel \: to \: XY \: and \: at \: a \: distance \: of \: 8 \: cm \: from \: A\\ \tt \:∠BAX + ∠AEC = 180° [Sum \: of \: co-interior \: angles \: is \: 180°]\\ \tt \:Now, ∠OAX =∠BAX= 90°\\ \tt \:[Tangent \: at \: any \: point \: of \: circle \: is \: perpendicular \: to \: radius \: through \: the \: point \: of \: contact]\\ \tt \:we \: have\\ \tt \:90 + ∠AEC = 180\\ \tt \:∠AEC = 90°\\ \tt \:Therefore, \:  ∠OEC = 90°\\ \tt \:So, \: OE ⏊ CD \: and  \: △OCB \: is \: right-angled \: triangle\\ \tt \ in △ \: OEB\\ \tt \:By \: Pythagoras \: theorem\\ \tt \:{ \red{ (base)2  + (perpendicular)2 =(hypotenuse)2 }}\\ \tt \:(OE)2 + (EC)2 = (OC)2\\ \tt \:And \: we \: have\\ \tt \:OC = 5 \: cm [radius]\\ \tt \:OE = AE - AO = 8 - 5 = 3 \: cm\\ \tt \:(3)2 + (EC)2= (5)2\\ \tt \:9 + (EC)2 = 25\\ \tt \:(EC)2= 25 - 9 = 16\\ \tt \:EC = 4 \: cm\\ \tt \:Also, \: CE = ED\\ \tt \:[Perpendicular \: from \: center \: to \: the \: chord \: bisects \: the \: chord] \\ \tt \: So, CD = CE + ED = 2CE = 2(4) = 8 cm$

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