eroshea Group Title 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. one year ago one year ago

1. hartnn Group Title

polygon of n sides have how many diagonals ?

2. eroshea Group Title

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

3. hartnn Group Title

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

4. eroshea Group Title

how?

5. hartnn Group Title

take side of one polygon as x and other as y x+y=13 [x(x-3)]/2+[y(y-3)]/2 = 25

6. eroshea Group Title

after that?

7. hartnn Group Title

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

8. eroshea Group Title

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

9. hartnn Group Title

yes! that is correct. u had options ?

10. eroshea Group Title

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?

11. hartnn Group Title

by option , i meant choices.

12. eroshea Group Title

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

13. hartnn Group Title

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

14. eroshea Group Title

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

15. pjaesecarilla Group Title

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

16. hartnn Group Title

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

17. hartnn Group Title

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

18. hartnn Group Title

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.

19. hartnn Group Title

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 ?

20. pjaesecarilla Group Title

21. hartnn Group Title

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

22. pjaesecarilla Group Title

ummm is xy=40?

23. hartnn Group Title

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

24. pjaesecarilla Group Title

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 :/

25. hartnn Group Title

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

26. hartnn Group Title

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

27. pjaesecarilla Group Title

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

28. imehs18 Group Title

29. imehs18 Group Title