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Q:

∆ABC में, ∠C=900, बिंदु P और Q क्रमशः AC और BC की भुजाओं पर हैं, इस प्रकार कि AP:PC=BQ:QC=1:2. फिर, $${AQ^2+BP^2}\over AB^2$$ बराबर है:

  • 1
    $$8\over 3$$
  • 2
    $$4\over 3$$
  • 3
    $$13\over 9$$
  • 4
    $$4\over 9$$
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Answer : 3. "$$13\over 9$$"

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