Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1182.
Thursday, January 28, 2016
Geometry Problem 1182: Two Squares, Four Quadrilaterals, Sum of the Areas, Metric Relations
Labels:
area,
metric relations,
quadrilateral,
square
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draw from E parallel to AD and to AB, at the same way from F, G, H
ReplyDeleteSideways EF, FG,GH,HE are formed 4 congr triangles, so and sideways
of a, b, c, d
And results is clear
http://s21.postimg.org/nszfowro7/pro_1182.png
ReplyDeleteDraw new square E’F’G’H’ as shown on the sketch
Let x= FF’=GG’=HH’=EE’
Define u, v, s, t as shown
Denote S(XYZ)= area of XYZ
Let S5= S(GFF’)=S(FEE’)=S(EHH’)=S(HGG’)
1. Note that u+v=s+t
With algebra calculation we get
.a^2 +c^2= u^2+v^2+s^2+t^2+2.x^2+2x(s+t)
And b^2+d^2= u^2+v^2+s^2+t^2+2.x^2+2x(u+v)
Since u+v= s+t => a^2 +c^2= b^2+d^2
2. Let S1’=S(BF’E’A), S2’=S(BF’G’C), S3’=S(CG’H’D) , S4’=S(AE’H’D)
Since u+v=s+t => S1’+S3’=S2’+S4’…. (1)
We have S1=S1’+S5+1/2x(t-s)
And S3=S3’+S5+1/2x(s-t)
So S1+S3=S1’+S3’+2.S5…. (2)
Similarly S2+S4=S2’+S4’+2.S5…(3)
Compare (1),(2) and (3) we will get S1+S3=S2+S4