Friday, September 18, 2009

Problem 356. Square, Interior point, Triangles, Area

Proposed Problem
Click the figure below to see the complete problem 356 about Square, Interior Point, Triangles, Area.

Problem 356. Square, Point inside, Triangles, Area.
See more:
Complete Problem 356
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:33 AM
Labels: , , ,

4 comments:

  1. yassin-NSeptember 21, 2009 at 10:51 AM

    we have:AB=BC=CD=AD=a
    S1+S3=(1/2*a*H1)+(1/2*a*H3)=1/2a*(H1+H3)=1/2*a*a
    S2+S4=(1/2*a*H2)+(1/2*a*H4)=1/2a*(H2+H4)=1/2*a*a
    then:S1+S3=S2+S4=1/2*a*a=1/2s.

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  2. AnonymousSeptember 22, 2009 at 9:23 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↓2+ S↓4=l(h↓2+ h↓4)/2= l↑2./2.
    QED, Ianuarius.

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  3. rv.littlemanNovember 10, 2020 at 1:24 AM

    See the drawing

    Define h1,h2,h3 and h4 the heights respectively of S1, S2, S3 and S4
    Define a the side of the square ABCD

    S1=a.h1/2
    S2=a.h2/2
    S3=a.h3/2
    S4=a.h4/2
    And h1+h3=h2+h4=a

    S1+S3=a(h1+h3)/2=a^2/2
    S2+S4=a(h2+h4)/2=a^2/2
    S=[ABCD]=a^2
    Therefore S1+S3= S2+S4=S/2

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  4. UnknownJanuary 24, 2021 at 8:48 AM

    Draw a line XPY parallel to AD, X on AB and Y on CD

    S2 = S(BCYX)/2 and S4 = S(ADYX)/2
    Adding S2 + S4 = S/2 = S1 + S3

    This result is true even if ABCD is a rectangle, a rhombus or a parallelogram

    Sumith Peiris
    Moratuwa
    Sri Lanka

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