త్రిభుజంలో ABC, AD, BE మరియు CF లు బిందువు G వద్ద కలిసే మధ్యస్థాలు మరియు  ABC త్రిభుజం వైశాల్యం 156 సెం.మీ2. FGE త్రిభుజం వైశాల్యం(సెం.మీ2లో) ఎంత?

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SSC CGL Previous Paper 11 (Held On: 10 August 2017 Shift 2)
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  1. 13
  2. 26
  3. 39
  4. 52

Answer (Detailed Solution Below)

Option 1 : 13
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ABC త్రిభుజం వైశాల్యం = 156 సెం.మీ2

F1 SSC Priya 9 10 24   D1

AD, BE మరియు CF అనేది బిందువు G వద్ద కలుస్తున్న మధ్యస్థాలు

కాబట్టి, D, E మరియు F బిందువులను కలిపే త్రిభుజ వైశాల్యం ABC త్రిభుజంలో 1/4వ వంతు అవుతుంది

DEF త్రిభుజం వైశాల్యం = (1/4) × 156 = 39 సెం.మీ2

ఇప్పుడు, G కూడా త్రిభుజం యొక్క కేంద్రకం అవుతుంది.

FGE త్రిభుజం వైశాల్యం = DFG త్రిభుజం వైశాల్యం =  DGE  త్రిభుజం వైశాల్యం= (1/3) × DEF త్రిభుజం వైశాల్యం

∴ FGE త్రిభుజం వైశాల్యం = (1/3) × 39 = 13సెం.మీ2
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