tag:blogger.com,1999:blog-6933544261975483399.post2048910413301635737..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1185: Similar Rectangles, Four Quadrilaterals, Sum of Areas, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-23536925931083062612016-02-04T09:33:34.848-08:002016-02-04T09:33:34.848-08:00Sum of S of trapezoids (' on AB,CD, " on ...Sum of S of trapezoids (' on AB,CD, " on AD, BC)<br />E"H"(AB-E'F'+AB-E'F')=? E'F'(AD-E"H"+AD-E"H")<br />E"H"AB-E"H"E'F' =? E'F'AD-E'F'E"H"<br />E"H"AB =? E'F'AD<br />AB/AD = E'F'/E"H" <br />c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59950518450972490732016-02-03T23:13:12.649-08:002016-02-03T23:13:12.649-08:00http://s25.postimg.org/5ag8xy18f/pro_1185_2.png
Se...http://s25.postimg.org/5ag8xy18f/pro_1185_2.png<br />See below for solution of problem 1185 part 2<br />From midpoint of each side of rectangle EFGH draw a new rectangle (red) with each side parallel to rectangle ABCD.<br />Define u’, v’, s’, t’ , z, w as shown on the sketch<br />Note that this red rectangle is similar to rect. ABCD and<br />(u’+v’)/(s’+t’)= AD/AB= z/w ….. (4)<br />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).<br /> Trapezoids areas of (S1+S3)= w(u’+v’)<br />And trapezoids areas of ( S2+S4).= z(s’+t’).<br />Using relation of (4) in above expressions we will get S1+S3=S2+S4<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50266101939551733752016-01-31T11:10:59.447-08:002016-01-31T11:10:59.447-08:00Second half of solution
draw EE' perpendicul...Second half of solution<br /> draw EE' perpendicular to AB and EE`` to AD.At the same way from F,G,H<br />Trapezoids EFF'E', GHH``G`` have altitude E'F' trapezoids FGF``G`` and EHH``E``<br />have altitude E``H``. Ratio E'F'/E``H``= AB/AD = EF/FH (similar tr with perp sides)<br />Sum of their bases are in ratio AD/AB=FH/EF (as diferences of inverse ratio)<br />First part of solutions is clear acording to pythaghor theorem included ratios<br />abovec.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26572636486500081472016-01-31T10:25:58.565-08:002016-01-31T10:25:58.565-08:00Draw new rectangle E’F’G’H’ similar to problem 118...Draw new rectangle E’F’G’H’ similar to problem 1182<br />Let k= AD/AB=EH/EF<br />Let x= FF’=HH’ =>EE’=GG’=k.x<br />All notations u, v,s, t , S(XYZ) will be the same as problem 1182<br />Perform length and width comparison of 2 similar rectangles ABCD and EFGH we will get<br />(u+v+x)/(t+s+k.x)= k …… (1)<br />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)<br />And b^+d^2= u^2+v^2+s^2+t^2+2x(x+u+v)…..(3)<br />Replace expression (1) in (2) we will get a^2+c^2=b^2+d^2<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com