ఒక సమద్విబాహు ΔMNP ఒక వృత్తంలో చెక్కబడి ఉంటుంది. MN = MP = \(16\sqrt{5}\) సెం.మీ, మరియు NP = 32 సెం.మీ అయితే, వృత్తం యొక్క వ్యాసార్థం (సెం.మీ.లో) ఎంత?

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SSC CGL 2021 Tier-I (Held On : 13 April 2022 Shift 3)
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  1. 18
  2. \(20\sqrt{5}\)
  3. \(18\sqrt{5}\)
  4. 20

Answer (Detailed Solution Below)

Option 4 : 20
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Detailed Solution

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ఇచ్చినది:

MN = MP = 16√5 సెం.మీ

NP = 32 సెం.మీ

ఉపయోగించిన సూత్రం:

త్రిభుజం వైశాల్యం = (1/2) x భూమి x ఎత్తు

పరివ్యాసార్ధము = ABC/4Δ

ఇక్కడ. A, B, మరియు C అనేవి త్రిభుజానికి మూడు భుజాలు

Δ = త్రిభుజం యొక్క వైశాల్యం

గణన:

F1 Madhuri SSC 22.06.2022 D16

MD NPకి లంబంగా ఉంటుంది

కాబట్టి, ND = DP = 32/2 = 16 సెం.మీ

MD = √[(16√5) 2 - (16) 2 ] = √(1280 - 256) = √1024 = 32 సెం.మీ.

ΔMNP వైశాల్యం = (1/2) x 32 x 32 = 512 సెం.మీ2

పరివ్యాసార్ధము = (16√5 x 16√5 x 32)/(4 x 512) = 20 సెం.మీ.

 ∴ వృత్తం యొక్క వ్యాసార్థం (సెం.మీ.లో) 20 సెం.మీ

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