anonymous
  • anonymous
Find the number of sides of each of the two polygons if the total number of sides of the polygons is 13, and the sum of the number of diagonals of the polygons is 25.
Mathematics
chestercat
  • chestercat
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hartnn
  • hartnn
polygon of n sides have how many diagonals ?
anonymous
  • anonymous
[n(n-3)]/2 ... is that it?
hartnn
  • hartnn
yes, so u could u form 2 equations using this formula?

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anonymous
  • anonymous
how?
hartnn
  • hartnn
take side of one polygon as x and other as y x+y=13 [x(x-3)]/2+[y(y-3)]/2 = 25
anonymous
  • anonymous
after that?
hartnn
  • hartnn
can u work out second equation and tell me what u get as x^2+y^2= ?
anonymous
  • anonymous
got it :)) the two polygon has 5 and 8 sides
hartnn
  • hartnn
yes! that is correct. u had options ?
anonymous
  • anonymous
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?
hartnn
  • hartnn
by option , i meant choices.
anonymous
  • anonymous
owh.. hahaha! my bad.. none :)
hartnn
  • hartnn
then u worked it out to get 5, 8 ? nice!
anonymous
  • anonymous
yes! :)) you helped me a lot there..
anonymous
  • anonymous
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
hartnn
  • hartnn
so, did u get how i got this : [x(x-3)]/2+[y(y-3)]/2 = 25 ?
hartnn
  • hartnn
when sides is 'x' number of diagonals is x (x-3)/2 same for y and total number of diagonals = 25
hartnn
  • hartnn
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.
hartnn
  • hartnn
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 ?
anonymous
  • anonymous
uhhhh help please :)
hartnn
  • hartnn
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
anonymous
  • anonymous
ummm is xy=40?
hartnn
  • hartnn
yes ! it is :) xy =40 is correct. what u got for x-y ?
anonymous
  • anonymous
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 :/
hartnn
  • hartnn
see, (x-y)^2 = x^2+y^2 -2xy = 89 -2(40) = 9 so, x-y = 3
hartnn
  • hartnn
x+y = 13 x-y =3 just add them
anonymous
  • anonymous
OOOOOOOOOOOOHHHHHHHHH!!!!!!! ok thanks man VERY MUCH haha i get it now :))))
anonymous
  • anonymous
hi pjaesecarilla and hartnn would like to ask about this question. im having problem as well.. how did you get the xy=40
anonymous
  • anonymous
hi @pjaesecarilla and @hartnn would like to ask about this question. im having problem as well.. how did you get the xy=40
hartnn
  • hartnn
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 :)
anonymous
  • anonymous
where did the xy come from? :) pls help thaaaaaaaaanks

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