Proposed Problem
Click the figure below to see the complete problem 357 about Square, Exterior point, Triangles, Area.
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Complete Problem 357
Level: High School, SAT Prep, College geometry
Friday, September 18, 2009
Problem 357. Square, Exterior point, Triangles, Area
Labels:
area,
exterior point,
square,
triangle
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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:
ReplyDeleteS↓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.
Draw EPF//BC, E on AB extended and F on DC extended
ReplyDeleteLet 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
See the drawing
ReplyDeleteDefine 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
Draw a Perpendicular from P equal to the length of side of the square (say h).
ReplyDeleteDenote 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