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420

Q:

In ∆ABC, ∠C=900, Point P and Q are on the sides AC and BC, respectively, such that  AP:PC=BQ:QC=1:2.Then, $${AQ^2+BP^2}\over AB^2$$  is equal to:

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

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