Thursday, October 9, 2008

Problem Of The Week 49

Find the maximum area that can be bound by four line segments of length 1, 2, 3 and 4. (you are not allowed to break a segment, you may join two :) )

1) 6√2

2) 2√6

3) 4√6

4) 3√2

5) None of these
Posted by Unknown at 11:45 AM
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9 comments:

  1. qwertyOctober 9, 2008 at 1:39 PM

    option e) none of these.
    can be triangle or quadrilateral.
    if triangle, by hero,s formula max. area is 2sqrt5
    and a quadrilateral cannot be formed of the given sides.

    ReplyDelete
  2. outtimedOctober 9, 2008 at 2:17 PM

    what made u think that quad can't be formed?

    ReplyDelete
  3. ankajOctober 9, 2008 at 2:47 PM

    We can arrange them in form of a trapezium, with sides 1 and 4 in parallel and two being the perpendicular distance between them...

    this gives area as 1/2* 2 * (1+4)=5...max area...correct me???

    ReplyDelete
  4. outtimedOctober 9, 2008 at 3:14 PM

    try again, are you able to make a trapezium?

    ReplyDelete
  5. ankajOctober 9, 2008 at 4:07 PM

    Yeah..i did a mistake didn't strike..trapezium is possible buy not with height as 2...it will be less than that...

    is the max area=2 sqrt(6)???

    ReplyDelete
  6. outtimedOctober 9, 2008 at 8:01 PM

    yeah , now it is fine

    ReplyDelete
  7. riyaOctober 11, 2008 at 11:12 PM

    Explain the answer please...

    ReplyDelete
  8. VarunOctober 12, 2008 at 4:36 AM

    me also got stumped by this questions ..
    well we can make quadilateral with sides 4,3,2,1 .. which has max area as root24
    trapezium too..

    ReplyDelete
  9. yodhaOctober 18, 2008 at 11:23 AM

    what clicked on my mind is that area is max. if all the vertices lie on a circle. so we can make a cyclic quadrilateral with the given lengths.
    and area will come out to be 2*sqrt(6) .

    ReplyDelete

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