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.

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) .

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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.

qwertyOctober 9, 2008 at 8:09 am

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

outtimedOctober 9, 2008 at 8:47 am

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???

ankajOctober 9, 2008 at 9:17 am

try again, are you able to make a trapezium?

outtimedOctober 9, 2008 at 9:44 am

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)???

ankajOctober 9, 2008 at 10:37 am

yeah , now it is fine

outtimedOctober 9, 2008 at 2:31 pm

Explain the answer please…

riyaOctober 11, 2008 at 5:42 pm

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..

VarunOctober 11, 2008 at 11:06 pm

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) .

yodhaOctober 18, 2008 at 5:53 am