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anonymous

  • one year ago

geometry problem

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  1. anonymous
    • one year ago
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    In triangle ABC, D, E and F are points on BC, AC and AB respectively such that BD: DC = 1:1, CE: EA = 1:2 and AF: FB = 1:3. If the area of triangle ABC = 48 cm2, find the area of triangle DEF.

  2. anonymous
    • one year ago
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    |dw:1433796310810:dw|

  3. MrNood
    • one year ago
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    I'm not sure that your diagram is correct You have used 'x' for BD, DC correctly, but I don't think you can use the 'same x' for AF and FB The diagram is correct but the nomenclature is wrong I think.

  4. anonymous
    • one year ago
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    Yeah think so but its in ratio it shoudnt matter much

  5. ganeshie8
    • one year ago
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    |dw:1433755799791:dw|

  6. MrNood
    • one year ago
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    @ganeshie8 There is no indication of HOW you arrived at those values Care to give an explanation instead of an unverified answer?

  7. ganeshie8
    • one year ago
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    Sure, thought id let others try too instead of spoiling the fun for them :)

  8. MrNood
    • one year ago
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    @aryandecoolest It does matter - if look at BDF your diagram implies BF is 3 times longer than BD, but that is not a correct interpretation of the question

  9. anonymous
    • one year ago
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    idk.. it's the exact question i have posted here... @ganeshie8 i didn't get... did we prove triangles similar or something

  10. MrNood
    • one year ago
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    Haven't had time to look in detail starting with area fro Side angle side a = absinC/2 and we have the same angle so we can compare ratios Looking at triangle BCA and triangle DCE DC/BC=1/2 CE/CA = 1/3 So the area of DCE = 1/6 ABC = 8 @ganeshie8 Not compatible with your answer

  11. perl
    • one year ago
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    Was that supposed to be x , not z ?

  12. perl
    • one year ago
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    @aryandecoolest Did the problem come with a picture, or did you draw that based on the directions?

  13. anonymous
    • one year ago
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    yeah it came with the picture....the same i have posted

  14. MrNood
    • one year ago
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    for bdf area = 1/2 * 3/4 = 3/8 ABC = 18 agreed for AFE Area = 1/4* 2/3 = 1/8 ABC = 8

  15. anonymous
    • one year ago
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    is there any theorem like area is equal to square of the proportional sides of 2 similar triangles?

  16. perl
    • one year ago
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    ok one moment :)

  17. MrNood
    • one year ago
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    @aryandecoolest use the area for 'side angle side' Area = absinC/2 apply that to the large triangle and the 3 corner triangles

  18. anonymous
    • one year ago
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    sine law ? but no angle given it's the ratios

  19. MrNood
    • one year ago
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    it's not sine law it's area from side angle side the ratios cancel the angles out

  20. anonymous
    • one year ago
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    Ah!!! fine... fine i get @MrNood

  21. Michele_Laino
    • one year ago
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    I think that this problem is equivalent to this one: "find three natural numbers, n_1, n_2, and n_3, such that the first one is disible by 2, the second one is divisible by 3, and the third one is divisible by 4": \[\begin{gathered} {n_1} = 2k \hfill \\ {n_2} = 3k \hfill \\ {n_3} = 4k \hfill \\ \end{gathered} \] nameli n_1, n_2 and n_3 are the sides of our triangle

  22. Michele_Laino
    • one year ago
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    furthermore, the subsequent condition holds: \[\begin{gathered} 2p = 9k \hfill \\ \frac{{9k}}{2} \times \frac{{5k}}{2} \times \frac{{3k}}{2} \times \frac{k}{2} = {48^2} \hfill \\ \end{gathered} \] where 2p is the perimeter of our triangle and the second formula comes from the application of Eron's formula

  23. Michele_Laino
    • one year ago
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    the solution is: \[\begin{gathered} {n_1} = 2k = \frac{8}{{\sqrt[4]{{15}}}} \hfill \\ {n_2} = 3k = \frac{{12}}{{\sqrt[4]{{15}}}} \hfill \\ {n_3} = 4k = \frac{{16}}{{\sqrt[4]{{15}}}} \hfill \\ \end{gathered} \]

  24. anonymous
    • one year ago
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    please give me a sec to understand....

  25. Michele_Laino
    • one year ago
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    now we have: \[48 = \frac{{{n_3}{n_1}}}{2}\sin \beta \] |dw:1433758100593:dw| from which: \[\sin \beta = \frac{3}{4}\sqrt {15} \]

  26. Michele_Laino
    • one year ago
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    finally we have: area of triangle FDB= \[ = \frac{{3{n_3}}}{4} \times \frac{{{n_1}}}{2} \times \sin \beta \times \frac{1}{2} = 18\] |dw:1433758258337:dw|

  27. Michele_Laino
    • one year ago
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    similarly for other triangles DCE, and AFE

  28. perl
    • one year ago
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    if you assume x= y , then you can use herons formula http://www.wolframalpha.com/input/?i=sqrt%28%284x%2B3x%2B2x%29%2F2+*%28%284x%2B3x%2B2x%29%2F2+-4x%29*%28%284x%2B3x%2B2x%29%2F2-3x%29*%28%284x%2B3x%2B2x%29%2F2-2x%29%29%3D48 I thought x and y were not necessarily equal

  29. Michele_Laino
    • one year ago
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    even if those numbers are irrational numbers

  30. anonymous
    • one year ago
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    okay thanks a ton @Michele_Laino @ganeshie8 @perl . I am still understanding... i'll contact u if i get a prob. Thanks a lot all of you. Hope i had more medals to give.

  31. Michele_Laino
    • one year ago
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    ok! @aryandecoolest

  32. MrNood
    • one year ago
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    there is NO indication that x and z (or y) are equal BD is NOT given as BF/3

  33. perl
    • one year ago
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    this problem is not well posed

  34. MrNood
    • one year ago
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    |dw:1433759138500:dw|

  35. perl
    • one year ago
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    The directions should say that x = y :)

  36. MrNood
    • one year ago
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    it is well posed - if you read the WORDS the problem is that somehow 'x' has been used to define the ratios of BC AND BA BF = 3z AF = z fits the words |dw:1433759367588:dw|

  37. perl
    • one year ago
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    @Michele_Laino using heron's I got k = 8/ 15^(1/4) therefore n1 = 2k = 16/ 15^(1/4) n2 = 3k = 24/ 15^(1/4) n3= 4k= 32/ 15^(1/4)

  38. MrNood
    • one year ago
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    @perl there is no given relationship between the sides this is ANY triangle

  39. ganeshie8
    • one year ago
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    |dw:1433759547800:dw|

  40. MrNood
    • one year ago
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    |dw:1433759530762:dw|

  41. perl
    • one year ago
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    @MrNood he stated that the diagram was given, z = x

  42. ganeshie8
    • one year ago
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    Ahh mistake, should be 8 and 14

  43. perl
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @perl @aryandecoolest Did the problem come with a picture, or did you draw that based on the directions? \(\color{blue}{\text{End of Quote}}\) \(\color{blue}{\text{Originally Posted by}}\) @aryandecoolest yeah it came with the picture....the same i have posted \(\color{blue}{\text{End of Quote}}\)

  44. MrNood
    • one year ago
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    It does not state that in the words

  45. perl
    • one year ago
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    Can you explain how you solved it without assuming x=y=z .

  46. MrNood
    • one year ago
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    yes I can the area of a triangle is given by Area = ab sinC/2

  47. MrNood
    • one year ago
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    so look at angle c for instance ABC area = (AC * CD sin C)/2 DCE area = (DC * CE * sin C)/2 so the ratio is the ratio of the product of the edges in this case theproduct is 1/6 so the area is 1/6 of ABC = 8 do the same for the two ther small triangles

  48. MrNood
    • one year ago
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    |dw:1433760746644:dw|

  49. MrNood
    • one year ago
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    |dw:1433760901106:dw|

  50. perl
    • one year ago
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    ok that makes sense. i just see a small typo

  51. perl
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @MrNood so look at angle c for instance ABC area = (AC * `CD` sin C)/2 that should be `CB` or `BC` DCE area = (DC * CE * sin C)/2 so the ratio is the ratio of the product of the edges in this case theproduct is 1/6 so the area is 1/6 of ABC = 8 do the same for the two ther small triangles \(\color{blue}{\text{End of Quote}}\)

  52. perl
    • one year ago
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    I'll just type out the argument and you can see if i have an error. CD = 1/2 BC CE = 1/3 AC ABC area = (AC * BC sin C)/2 = 48 DCE area = (CD * CE * sin C)/2 = ( 1/2 BC * 1/3 AC ) * sin C / 2 = 1/2 * 1/3 ( BC * AC) * sin C /2 = 1/6 * 48 = 8

  53. MrNood
    • one year ago
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    yes - that is correct - my apologies

  54. perl
    • one year ago
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    we can make a similar argument for angle B

  55. MrNood
    • one year ago
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    as I said - do the samer for the other small triangles |dw:1433761604217:dw|

  56. perl
    • one year ago
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    then subtract from 48 to get the middle area

  57. perl
    • one year ago
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    interestingly if you do happen to assume x=y=z, you still get the same answer

  58. perl
    • one year ago
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    but you shouldn't since it is not stated in the directions

  59. perl
    • one year ago
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    and if you assume x=y=z then the problem can be solved using Heron's formula. do you have any questions @aryandecoolest

  60. anonymous
    • one year ago
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    no, it's getting clear. thanks a lot

  61. MrNood
    • one year ago
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    @perl The solution I gave above is a general solution for all triangles It is not therefore 'interesting' that it works for the 'special case' x=y=z'

  62. Michele_Laino
    • one year ago
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    Yes! you are right! @perl

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