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Problem Of The Week 49

with 9 comments

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


Written by Implex

October 9, 2008 at 6:15 am

9 Responses

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  1. 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.


    October 9, 2008 at 8:09 am

  2. what made u think that quad can’t be formed?


    October 9, 2008 at 8:47 am

  3. 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???


    October 9, 2008 at 9:17 am

  4. try again, are you able to make a trapezium?


    October 9, 2008 at 9:44 am

  5. 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)???


    October 9, 2008 at 10:37 am

  6. yeah , now it is fine


    October 9, 2008 at 2:31 pm

  7. Explain the answer please…


    October 11, 2008 at 5:42 pm

  8. 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..


    October 11, 2008 at 11:06 pm

  9. 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) .


    October 18, 2008 at 5:53 am

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