Saturday, 22 October 2011

Geometry -1 (IIFT-2008)

If D is the mid point of side BC of a triangle ABC and AD is the perpendicular to AC then:
(A)  3AC2 = BC2-AB2
(B)  3BC2 = AC2-3AB2
(C)  BC2+ AC2 = 5AB2
(D) None of the above
To enter your answer, click on "comments" below:
Posted by at
Labels:

3 comments:

  1. Rohan15 April 2012 at 17:11

    A

    ReplyDelete
  2. arun17 August 2013 at 17:13

    D

    see consider a triangle with base 8cm(BC) and altitude 3(let be AD).


    since D bisects and form right angle,we have 2 RIGHT ANGLED TRIANGLE with height AD(=3cm)
    bases BD(=4cm) and DC(=4cm).

    then using pythogorus theorem,we get AB=5cm and AD=5cm
    subtituting the values in the above four options ,i am getting none of them satisfied

    SO answer option is D)NONE of the above

    ReplyDelete
    Replies
    1. MathByVemuri17 August 2013 at 22:18

      In the question, it is given that AD is perpendicular to AC but not to BC...

      Delete

Subscribe to: Post Comments (Atom)