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polygon of n sides have how many diagonals ?
[n(n-3)]/2 ... is that it?
yes, so u could u form 2 equations using this formula?
take side of one polygon as x and other as y x+y=13 [x(x-3)]/2+[y(y-3)]/2 = 25
can u work out second equation and tell me what u get as x^2+y^2= ?
got it :)) the two polygon has 5 and 8 sides
yes! that is correct. u had options ?
i guess none... im just finding the sides of the two polygons is there a possible option for this problem? if there is, can you tell me what it may be?
by option , i meant choices.
owh.. hahaha! my bad.. none :)
then u worked it out to get 5, 8 ? nice!
yes! :)) you helped me a lot there..
hello there :) can i ask @hartnn what you meant by "can u work out second equation and tell me what u get as x^2+y^2= ?" sorry i also encountered this problem on my solid mensuration
so, did u get how i got this : [x(x-3)]/2+[y(y-3)]/2 = 25 ?
when sides is 'x' number of diagonals is x (x-3)/2 same for y and total number of diagonals = 25
that can be simplified as \([x(x-3)]/2+[y(y-3)]/2 = 25 \\ x^2-3x+y^2-3y=50 \\ (x^2+y^2)-3(x+y)=50\) but we know x+y =13 so, \(x^2+y^2-3\times 13=50\) we can easily find x^2+y^2 from here.
and once we know x^2+y^2 and x+y , we can find x and y you know how to ? or need help with that ?
uhhhh help please :)
ok, so we use the fact that (x+y)^2 = x^2+y^2 +2xy from this we will get xy. right ? then (x-y)^2 = x^2+y^2 -2xy from this we will get x-y , ok ? and now we have x-y, x+y if we just add then, 'y' gets eliminated and you get the value of x try once, if you get stuck, i will show you the entire solution
ummm is xy=40?
yes ! it is :) xy =40 is correct. what u got for x-y ?
i kinda got stuck on that one. (x-y)^2 = x^2+y^2-2xy x^2=89-y^2 so. 89-y^2 + y^2 - 2(40) so i got 9. i dunno if im correct :/
see, (x-y)^2 = x^2+y^2 -2xy = 89 -2(40) = 9 so, x-y = 3
x+y = 13 x-y =3 just add them
OOOOOOOOOOOOHHHHHHHHH!!!!!!! ok thanks man VERY MUCH haha i get it now :))))
hi pjaesecarilla and hartnn would like to ask about this question. im having problem as well.. how did you get the xy=40
hi @pjaesecarilla and @hartnn would like to ask about this question. im having problem as well.. how did you get the xy=40
x+y = 13 x^2+y^2 -39 = 50 , so x^2+y^2 = 50+39 = 89 (x+y)^2 = x^2+2xy +y^2 13^2 = 89 + 2xy 169 = 89 +2xy 2xy =169-89 = 80 xy =40 :)
where did the xy come from? :) pls help thaaaaaaaaanks