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  • Question

  • Consider the following statements: 1) The perimeter of a triangle is greater than the sum of its three medinas. 2) In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD. Which of the above statements is/are correct?


  • Options
  • A. 1 only
  • B. 2 only
  • C. Both 1 and 2
  • D. Neither 1 nor 2

  • Correct Answer
  • Both 1 and 2 

    Explanation
    Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
    Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)
    Hence in ?ABD, AD is a median
    ?
    AB + AC > 2(AD)
    Similarly, we get
    BC + AC > 2CF
    BC + AB > 2BE
    On adding the above inequations, we get
    (AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
    2(AB + BC + AC) > 2(AD + BE + CF)
    ?
    AB + BC + AC > AD + BE +CF
     
    2.
    To prove: AB + BC + CA > 2AD
    Construction: AD is joined
    Proof: In triangle ABD,
    AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the
    third side]
    ----
    1
    In triangle ADC,
    AC + DC > AD [because, the sum of any two
    sides of a tri
    angle is always greater than the
    third side]
    ----
    2
    Adding 1 and 2 we get,
    AB + BD + AC + DC > AD + AD
    => AB + (BD + DC) + AC > 2AD
    => AB + BC + AC > 2AD
    Hence proved
    • Tags: Bank Exams

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