1. Subscriberjoe shmo
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    29 Apr '16 22:43
    1) Triangle "ABC" is a right triangle with hypotenuse "AC" of 60.
    2) Point "E" lies on line "BC" such that right triangle "ABE" has a hypotenuse "AE" of 52.
    3) Point "D" lies on line "AB" such that right triangle "DBC" has a hypotenuse "DC" of 39.

    Find Length "DE"?
  2. Standard memberwolfgang59
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    01 May '16 09:19
    Originally posted by joe shmo
    1) Triangle "ABC" is a right triangle with hypotenuse "AC" of 60.
    2) Point "E" lies on line "BC" such that right triangle "ABE" has a hypotenuse "AE" of 52.
    3) Point "D" lies on line "AB" such that right triangle "DBC" has a hypotenuse "DC" of 39.

    Find Length "DE"?
    mmmm
    Deceptively tricky.
    I have 5 unknowns and only 4 equations.
    Must be a little geometric trick in there that I have forgotten.
    🙂
  3. Standard memberwolfgang59
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    01 May '16 09:271 edit
    But that's ok.
    ("DE" )^2 = (52*52)-(21*99)
  4. Subscriberjoe shmo
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    01 May '16 15:092 edits
    Originally posted by wolfgang59
    But that's ok.
    ("DE" )^2 = (52*52)-(21*99)
    Yep, you got it! It is certainly more tricky than it first appears.

    Let:
    AB = A
    BC = B
    AD = a
    EC = b

    A^2+B^2 = 60^2 (1)

    A^2 + (B-b)^2 = 52^2 (2)

    (A-a)^2 + B^2 = 39^2 (3)

    Let:

    (A-a) = F
    (B-b) = G

    Substitute F and G into (2)&(3)

    System:

    A^2 + B^2 = 60^2 (1)

    A^2 + G^2 = 52^2 ( 2' )

    F^2 + B^2 = 39^2 ( 3' )

    ( 2' )+( 3' ) :

    A^2 + G^2 + F^2 + B^2 = 52^2 + 39^2 (4)

    (4) - (1):

    A^2 + G^2 + F^2 + B^2 -A^2 -B^2 = 52^2 + 39^2-60^2

    G^2+F^2 = 52^2 + 39^2-60^2 = 625

    DE = Sqrt(G^2+F^2) = 25
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