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.

## Saturday, January 30, 2016

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

Labels:
area,
metric relations,
rectangle,
similarity

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Draw new rectangle E’F’G’H’ similar to problem 1182

ReplyDeleteLet k= AD/AB=EH/EF

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

http://s25.postimg.org/5ag8xy18f/pro_1185_2.png

DeleteSee 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

(u’+v’)/(s’+t’)= AD/AB= z/w ….. (4)

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

Second half of solution

ReplyDeletedraw 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

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

ReplyDeleteE"H"(AB-E'F'+AB-E'F')=? E'F'(AD-E"H"+AD-E"H")

E"H"AB-E"H"E'F' =? E'F'AD-E'F'E"H"

E"H"AB =? E'F'AD

AB/AD = E'F'/E"H"