Friday, May 21, 2010

Problem 459: Right triangle, Squares, Distance, Measurement

Geometry Problem
Click the figure below to see the complete problem 459 about Right triangle, Squares, Distance, Measurement.

Problem 459: Right triangle, Squares, Distance, Measurement.
See also:
Complete Problem 459

Level: High School, SAT Prep, College geometry

5 comments:

  1. cos(GAD)= cos(GAC+CAD)=cos(GAC+45)=cos(GAC)cos(45)-sin(GAC)sin(45). But cos(GAC)=c/b & sin(GAC)=a/b hence cos(GAD)=(c-a)/(b√2). From Tr. GAD using the co-sine rule, we’ve: x^2 = GA^2 +AD^2 -2GA*AD*cos(GAD) = (c+a)^2+2b^2-2*(c+a)( b√2)(c-a)/(b√2)= (c+a)^2 + 2(c^2+a^2) -2*(c+a)(c-a)= c^2 + 2ac + a^2 + 4a^2 = 5a^2 + c^2 + 2ac
    Ajit

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  2. Let m projection of AB over AC.
    Let n projection of BC over AC.
    Let h altitude over hypotenuse AC.
    Then (T.altitude) h = sqrt (mn)
    Then (T.legs) m = c^2/sqrt(a^2+c^2)
    n = a^2/sqrt(a^2+c^2)
    Let H point of intersection of parallel to AC from G and extension of side DC from C. Then GHD is right triangle on H.
    We've the next relations:
    CD = AC = m + n = sqrt(a^2 +c^2) (Pt. Tri.ABC)
    CH = h + n
    GH = h - n
    HD = CH + CD = h + 2n + m
    Then:
    x^2=GH^2 + HD^2 = (h-n)^2 + (h + 2n + m)^2 =
    = ... = 6mn + 2*sqrt(mn)*(m+n) + 5n^2 + m^2 = ... =
    = 5a^2 + 2ac + c^2

    MIGUE.

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  3. Let H point on line GF, and HF=GF. Then CG=CF and angle GCH=90.

    On triangle CDG and triangle CAH, CD=CA and CG=CH.
    angle DCG=90+angle DCH, angle ACH=90+angle DCH.

    So, triangle CDG==triangle CAH ---> then, AH=DG=x

    On triangle AGH, angle G=90, AG=a+c, GH=2a, AH=x.

    Therefore, x^2=(a+c)^2+(2a)^2 ---> x^2 = 5a^2+c^2+2ac

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  4. @ third comment by Anonymous:
    CG will never be equal to CF as CG is the diagonal of the square and CF the side
    Pls correct this comment

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  5. By boxing the largest square ACDE its possible to make a new right triangle with x as its hypotenuse that's easy to compute:

    First rotate and translate triangle ABC so that its hypotenuse lies on CD and its new vertex is H.
    Then extend the DH side so that it meets FG at I forming a new right triangle.

    DI = a + a
    GI = a + c
    x^2 = DI^2 + GI^2 = (2a)^2 (a+c)^2 = 5a^2 + 2ac + c^2

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