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DLS

  • 3 years ago

An interesting question! :D ABC is an equilateral triangle such that the vertices B and C lie on two parallel line at a distance of 6.If A lies between the parallel lines at a distance 4 from one of them then the length of a side of the equilateral triangle is=?

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  1. hartnn
    • 3 years ago
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    if B and C lie on parallel lines, then BC = side = 6 ...?

  2. DLS
    • 3 years ago
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    units

  3. DLS
    • 3 years ago
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    no unit mentioned if u are asking for one

  4. hartnn
    • 3 years ago
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    i was not asking, i was telling side =6

  5. DLS
    • 3 years ago
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    oh :P

  6. DLS
    • 3 years ago
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    |dw:1356782149154:dw| my diagram

  7. hartnn
    • 3 years ago
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    if ABC is equilateral AB=BC=AC=6

  8. DLS
    • 3 years ago
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    |dw:1356782275365:dw| so cant we do like this.. sin60=h/4

  9. hartnn
    • 3 years ago
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    u wanted to know length of side ... = BC = 6 thats it.

  10. DLS
    • 3 years ago
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    O_o answer isnt this!!

  11. DLS
    • 3 years ago
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    we have to find length

  12. hartnn
    • 3 years ago
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    of ?

  13. DLS
    • 3 years ago
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    side of the eq triangle :/

  14. hartnn
    • 3 years ago
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    maybe the side BC is tilted , then

  15. DLS
    • 3 years ago
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    yes..thats what im saying

  16. hartnn
    • 3 years ago
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    so, your figure is incorrect.

  17. DLS
    • 3 years ago
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    :/

  18. hartnn
    • 3 years ago
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    where u have shown BC =6

  19. experimentX
    • 3 years ago
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    |dw:1356782465079:dw|

  20. DLS
    • 3 years ago
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    why wont that be 60 :o

  21. hartnn
    • 3 years ago
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    |dw:1356782731705:dw|

  22. DLS
    • 3 years ago
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    thats nt touching?

  23. hartnn
    • 3 years ago
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    ofcourse, it is...

  24. DLS
    • 3 years ago
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    then w|dw:1356783287510:dw|

  25. hartnn
    • 3 years ago
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    i have a long way.

  26. hartnn
    • 3 years ago
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    |dw:1356783439425:dw|

  27. hartnn
    • 3 years ago
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    \(a=\sqrt{x^2-4^2} \\ b=\sqrt{x^2-6^2} \\ a+b=\sqrt{x^2-2^2}\) you can find 'x' from here.

  28. hartnn
    • 3 years ago
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    i hope ther's a shorter way...

  29. DLS
    • 3 years ago
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    x=root 52 or smt?

  30. DLS
    • 3 years ago
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    solution and answer given in my book is DANGEROUS :/

  31. hartnn
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=sqrt%28x%5E2-16%29%2Bsqrt%28x%5E2-36%29%3Dsqrt%28x%5E2-4%29 what answer u have ?

  32. DLS
    • 3 years ago
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    I have.. \[\LARGE 2 \frac{\sqrt{28}}{\sqrt{3}}\]

  33. hartnn
    • 3 years ago
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    which is exactly what i got.

  34. hartnn
    • 3 years ago
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    \(\LARGE 2 \frac{\sqrt{28}}{\sqrt{3}}=\LARGE 4\frac{\sqrt{7}}{\sqrt{3}}\)

  35. DLS
    • 3 years ago
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    okay yes

  36. hartnn
    • 3 years ago
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    using \(a=\sqrt{x^2-4^2} \\ b=\sqrt{x^2-6^2} \\ a+b=\sqrt{x^2-2^2}\)

  37. DLS
    • 3 years ago
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    hold on one sec..let me go through it once again

  38. DLS
    • 3 years ago
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    lol that was a nice method :|

  39. hartnn
    • 3 years ago
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    just pythagoras, thrice...

  40. DLS
    • 3 years ago
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    yeah..!

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