## Saturday, January 30, 2016

### Geometry Problem 1185: Similar Rectangles, Four Quadrilaterals, Sum of Areas, Metric Relations

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 1185.

1. Draw new rectangle E’F’G’H’ similar to problem 1182
Let x= FF’=HH’ =>EE’=GG’=k.x
All notations u, v,s, t , S(XYZ) will be the same as problem 1182
Perform length and width comparison of 2 similar rectangles ABCD and EFGH we will get
(u+v+x)/(t+s+k.x)= k …… (1)
Similar to pro. 1182 we have a^2+c^2= u^2+v^2+s^2+t^2+2k.x(k.x+t+s)…..(2)
And b^+d^2= u^2+v^2+s^2+t^2+2x(x+u+v)…..(3)
Replace expression (1) in (2) we will get a^2+c^2=b^2+d^2

1. http://s25.postimg.org/5ag8xy18f/pro_1185_2.png
See below for solution of problem 1185 part 2
From midpoint of each side of rectangle EFGH draw a new rectangle (red) with each side parallel to rectangle ABCD.
Define u’, v’, s’, t’ , z, w as shown on the sketch
Note that this red rectangle is similar to rect. ABCD and
At each corner of ABCD we have 2 equal areas triangles so the problem become to show that trapezoids areas of (S1+S3)= trapezoids areas of ( S2+S4).
Trapezoids areas of (S1+S3)= w(u’+v’)
And trapezoids areas of ( S2+S4).= z(s’+t’).
Using relation of (4) in above expressions we will get S1+S3=S2+S4

2. Second half of solution
draw EE' perpendicular to AB and EE`` to AD.At the same way from F,G,H
Trapezoids EFF'E', GHH``G`` have altitude E'F' trapezoids FGF``G`` and EHH``E``
have altitude E``H``. Ratio E'F'/E``H``= AB/AD = EF/FH (similar tr with perp sides)
Sum of their bases are in ratio AD/AB=FH/EF (as diferences of inverse ratio)
First part of solutions is clear acording to pythaghor theorem included ratios
above

3. Sum of S of trapezoids (' on AB,CD, " on AD, BC)