Monday, August 25, 2008

Elearn Geometry Problem 163



See complete Problem 163 at:
www.gogeometry.com/problem/p163_trapezoid_triangle_area.htm

Trapezoid, Diagonals, Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 11:25 AM
Labels: , , , , ,

7 comments:

  1. AjitApril 16, 2009 at 8:31 PM

    Let A be(0,0), B:(b.h),C:(c,h) & D:(d,0)
    It can easily be seen that triangles ABC & DBC have the base and height and hence, Tr ABC =Tr. DBC or S3 + S1 = S4 + S1 or S3 = S4.
    Now height of E above AD can be determined as hd/(-b+c+d) and thus, S2 = hd^2/2(-b+c+d) while S1=(c-b)/2 *[h - hd/(-b+c+d))=h(c-b)^2/2(-b+c+d) which gives us S1*S2 =(hd(c-b))^2/4(-b+c+d)^2-(1)
    While S3 =(1/2)[cdh/(-b+c+d)- bhd/(-b+c+d)]
    =(1/2)hd(c-b)/(-b+c+d) -----------(2)
    By (1) & (2) we've, S3^2=S1*S2 or S3 = V(S1*S2) where V=square root.
    Now S=S1+S2+S3+S4= S1+S2+2S3 = S1+S2+2*V(S1*S2) Hence, VS = VS1 + VS2
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. AnonymousSeptember 23, 2009 at 1:02 PM

    By JCA
    2)
    S=S1+S2+S3+S4=S1+S2+2sqrt(S1*S2)=[sqrt(S1)+sqrt(S2)]^2
    hence
    sqrt(S)=sqrt(S1)+sqrt(S2)

    ReplyDelete
  3. ulksterFebruary 20, 2012 at 3:00 AM

    Could you explain how you got E's height in a little more detail? I like the proof otherwise.

    ReplyDelete
  4. PravinFebruary 20, 2012 at 4:19 AM

    E = AC ∩ BD
    Solve for y the equations
    hx - cy = o (equation of AC),
    hx - (b - d)y - hd = 0 (equation of BD)
    at E:(b-d -c)y + hd = 0
    or y = hd /(c + d - b)

    ReplyDelete
  5. UnknownApril 27, 2012 at 9:47 AM

    The solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJV1U1OFlBMjhuN2s

    ReplyDelete
  6. Nilton LapaJuly 10, 2012 at 4:33 PM

    Solution to problem 163.
    1) Triangles ABC and BCD have the same basis BC, and equal altitudes, so they have the same area. Hence
    S3 = S(ABC) – S(BCE) = S(BCD) – S(BCE) = S4.
    Triangles ABE and EBC have the same altitude relative to line AC. Then S3/S1 = AE/EC.
    Triangles ABE and AED have the same altitude relative to line BD. Then S2/S3 = DE/EB.
    But AED and CEB are similar and AE/EC = DE/EB. Then S3/S1 = S2/S3 and S3 = S4 = sqrt(S1.S2).

    2) We have
    S = S1 + S2 + S3 + S4 =
    = S1 + S2 + 2.sqrt(S1.S2) =
    = (sqrt(S1) + sqrt(S2))^2
    Then sqrt(S) = sqrt(S1) + sqrt(S2).

    ReplyDelete
  7. MarcoFebruary 3, 2023 at 6:59 AM

    Triangle BCE ~ Triangle DAE
    Let BC:AD=1:k, so height of triangle BCE : height of triangle DAE also =1:k
    Let BC=a, AD=ka, height of triangle BCE=b and height of triangle DAE =kb
    (1)S1+S3=S1+S4(same base same height)
    S3=S4=ab(1+k)/2-ab/2=abk/2
    (S1)(S2)=(ab/2)*(abk^2)/2=(abk/2)^2
    So S3=S4=sqrt((S1)(S2))

    (2)S=(a+ka)(b+bk)/2=[ab(1+k)^2]/2
    sqrt(S)=(1+k)*sqrt(ab/2)
    From(1), sqrt(S1)=sqrt(ab/2), sqrt(S2)=k*sqrt(ab/2)
    By simple calculation, sqty(S)=sqrt(S1)+sqrt(S2)

    ReplyDelete

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