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.

- anonymous

- katieb

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- hartnn

polygon of n sides have how many diagonals ?

- anonymous

[n(n-3)]/2 ... is that it?

- hartnn

yes, so u could u form 2 equations using this formula?

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## More answers

- anonymous

how?

- 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

after that?

- hartnn

can u work out second equation and tell me what u get as x^2+y^2= ?

- anonymous

got it :)) the two polygon has 5 and 8 sides

- hartnn

yes! that is correct.
u had options ?

- 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

by option , i meant choices.

- anonymous

owh.. hahaha! my bad.. none :)

- hartnn

then u worked it out to get 5, 8 ? nice!

- anonymous

yes! :)) you helped me a lot there..

- 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

so, did u get how i got this :
[x(x-3)]/2+[y(y-3)]/2 = 25
?

- hartnn

when sides is 'x' number of diagonals is x (x-3)/2
same for y
and total number of diagonals = 25

- 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

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

uhhhh help please :)

- 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

ummm is xy=40?

- hartnn

yes ! it is :)
xy =40 is correct.
what u got for x-y ?

- 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

see, (x-y)^2 = x^2+y^2 -2xy = 89 -2(40) = 9
so, x-y = 3

- hartnn

x+y = 13
x-y =3
just add them

- anonymous

OOOOOOOOOOOOHHHHHHHHH!!!!!!! ok thanks man VERY MUCH haha i get it now :))))

- 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

hi @pjaesecarilla and @hartnn would like to ask about this question. im having problem as well.. how did you get the xy=40

- 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

where did the xy come from? :) pls help thaaaaaaaaanks

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