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eroshea

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

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  1. hartnn
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    polygon of n sides have how many diagonals ?

    • one year ago
  2. eroshea
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    [n(n-3)]/2 ... is that it?

    • one year ago
  3. hartnn
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    yes, so u could u form 2 equations using this formula?

    • one year ago
  4. eroshea
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    how?

    • one year ago
  5. hartnn
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    take side of one polygon as x and other as y x+y=13 [x(x-3)]/2+[y(y-3)]/2 = 25

    • one year ago
  6. eroshea
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    after that?

    • one year ago
  7. hartnn
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    can u work out second equation and tell me what u get as x^2+y^2= ?

    • one year ago
  8. eroshea
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    got it :)) the two polygon has 5 and 8 sides

    • one year ago
  9. hartnn
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    yes! that is correct. u had options ?

    • one year ago
  10. eroshea
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    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?

    • one year ago
  11. hartnn
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    by option , i meant choices.

    • one year ago
  12. eroshea
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    owh.. hahaha! my bad.. none :)

    • one year ago
  13. hartnn
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    then u worked it out to get 5, 8 ? nice!

    • one year ago
  14. eroshea
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    yes! :)) you helped me a lot there..

    • one year ago
  15. pjaesecarilla
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    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

    • 5 months ago
  16. hartnn
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    so, did u get how i got this : [x(x-3)]/2+[y(y-3)]/2 = 25 ?

    • 5 months ago
  17. hartnn
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    when sides is 'x' number of diagonals is x (x-3)/2 same for y and total number of diagonals = 25

    • 5 months ago
  18. hartnn
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    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.

    • 5 months ago
  19. hartnn
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    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 ?

    • 5 months ago
  20. pjaesecarilla
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    uhhhh help please :)

    • 5 months ago
  21. hartnn
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    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

    • 5 months ago
  22. pjaesecarilla
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    ummm is xy=40?

    • 5 months ago
  23. hartnn
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    yes ! it is :) xy =40 is correct. what u got for x-y ?

    • 5 months ago
  24. pjaesecarilla
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    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 :/

    • 5 months ago
  25. hartnn
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    see, (x-y)^2 = x^2+y^2 -2xy = 89 -2(40) = 9 so, x-y = 3

    • 5 months ago
  26. hartnn
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    x+y = 13 x-y =3 just add them

    • 5 months ago
  27. pjaesecarilla
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    OOOOOOOOOOOOHHHHHHHHH!!!!!!! ok thanks man VERY MUCH haha i get it now :))))

    • 5 months ago
  28. imehs18
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    hi pjaesecarilla and hartnn would like to ask about this question. im having problem as well.. how did you get the xy=40

    • 4 months ago
  29. imehs18
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    hi @pjaesecarilla and @hartnn would like to ask about this question. im having problem as well.. how did you get the xy=40

    • 4 months ago
  30. hartnn
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    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 :)

    • 4 months ago
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