wat i did to solve this question during the test was.....i divided the 2 diagonal into two parts each-lets say a a' b and b'
the then formed equations with pythagoras theorem.....three equations....compared it with the options to see which options suit the equations...from the equations i saw dat the square of the 4th side must be 5 less then or 5 more than 81.....so hence ticked root 76 and root 86....maybe my approach is not mathematically correct but it worked for me......
A quadrilateral has sides 2,3,9....diagonals are at rt.angles.
Find all possible values of the 4th side....
-
UP 0 DOWN 0 0 17
17 Answers
hey..nishant bhaiya..its Avinav Prakash 4m Heritage....wat ive is that ...taken just x as the lone variable...n used pytha..3 small steps and 2 diff cases..and ans was √76 and √86.....
guys this one is not that calculation intensive...
Just revisit what i have written in #3.
Or else i will show you what can be done.
Yup ronald, I know dat, but still I feel raja is r8.
As it is u have it not short enuf after eleminating those 4 variables - now to do it again with a new set of sides is worse than horrible....
Moreover Nishant bhaiya's method might make it shorter.
@ raja : this is fiitjee aits,,karna padega.
@ avinav : ur probably right.
array sides change karkey dekho.....thodey aur ans aa jayengey
i mean to say wen sides with length (2 and 9 )are adjacent ....wen (9 and 3) are adjacent and wen (2 and 3) r adjacent
aur bhi answers they kya?????[11][12]
is this a fiitjee aits q?
ok i think i got it......teh diagonals of the quadrilateral divide teh quadrilateral into 4 right angle traingles
so from pythagoras theo.
t2+u2=4....................i
w2+v2=81.................ii
u2+v2=9....................iii
i +ii - iii
we get
t2+w2=4+81-9=76
t2+w2=x2=76
x=√76
Wasnt it easy [3]
array bhaiya is hint sey kuch kaam nai aaya :P...
that thing was in my mind......how to proceed after that ??
Think of this one geometrically.
First show (convince yourself) that 2 and 3 have to be the adjacent sides...
Then geometrically see how much you can slide the 2 sides 2 and 3 along the diagonal....
I hope this hint gives you a starting point [1]