Question

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.

Answer 1

polygon of n sides have how many diagonals ?

Answer 2

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

Answer 3

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

Answer 4

how?

Answer 5

take side of one polygon as x and other as y
x+y=13

[x(x-3)]/2+[y(y-3)]/2 = 25

Answer 6

after that?

Answer 7

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

Answer 8

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

Answer 9

yes! that is correct.
u had options ?

Answer 10

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?

Answer 11

by option , i meant choices.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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.

Answer 19

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 ?

Answer 20

uhhhh help please :)

Answer 21

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

Answer 22

ummm is xy=40?

Answer 23

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

Answer 24

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 :/

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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
:)

Answer 31

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