Friday, September 18, 2009

Problem 357. Square, Exterior point, Triangles, Area

Proposed Problem
Click the figure below to see the complete problem 357 about Square, Exterior point, Triangles, Area.

Problem 357. Square, Exterior point, Triangles, Area.
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Complete Problem 357
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:57 AM
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4 comments:

  1. AnonymousSeptember 22, 2009 at 9:25 AM

    Let l be the side of the square ABCD and h1, h2, h3, h4 be respectively the heights of triangles PAB, PBC, PCD, PDA. We have:
    S↓1=lh↓1/2, S↓2 = lh↓2/2, S↓3=l h↓3/2, S↓4=l h↓4/2. Therefore:

    S↓1+ S↓3=l(h↓1+ h↓3)/2= l↑2./2,
    S↓4- S↓2=l(h↓4- h↓2)/2= l↑2./2.
    QED, Ianuarius.

    ReplyDelete
  2. Sumith PeirisJune 18, 2016 at 12:47 AM

    Draw EPF//BC, E on AB extended and F on DC extended

    Let EP = x and EB = y

    So S1 = xa/2, S3 = (a-x)a/2,
    S2 = ya/2 and S4 = (a+y)a/2

    So S1+S3 = S4-S2= a^2/2 = S/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. rv.littlemanNovember 10, 2020 at 5:07 AM

    See the drawing

    Define a = side of square ABCD
    Define h1,h2,h3 and h4 the heights respectively of S1, S2, S3 and S4
    h1+h3=a, h4-h2=a
    S=[ABCD]=a^2
    2S1=a.h1, 2S3=a.h3
    2S1+2S3=a(h1+h3)=a^2
    2S2=a.h2, 2S4=a.h4
    2S4-2S2=a(h4-h2)=a^2
    Therefore S1+S3=S4-S2=S/2

    ReplyDelete
  4. Sailendra TNovember 19, 2020 at 10:21 PM

    Draw a Perpendicular from P equal to the length of side of the square (say h).
    Denote it as O.Connect AO & DO to form two parallelograms ABPO,DCPO and a triangle AOD congruent to BPC.
    It can easily seen that S1+S2+S3=S4
    Hence S1+S3= S4-S2
    Let the length of Perpendicular from P to BC be h1
    => S4-S2 = 0.5(h1.h+h2-h1.h)=h^2/2=S/2

    ReplyDelete

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