tag:blogger.com,1999:blog-6933544261975483399.post4582230305210616386..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 239: Square, Midpoints, Congruence, PythagorasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-6933544261975483399.post-74880093181447119352023-03-12T07:05:59.951-07:002023-03-12T07:05:59.951-07:00Triangle BME ~ Triangle BCF
BM:ME=2:1
ME=c, BM=2c
...Triangle BME ~ Triangle BCF<br />BM:ME=2:1<br />ME=c, BM=2c<br />----------------------------------------<br />By Pyth. Theorem (Triangle BME)<br />BE^2=BM^2+ME^2=5c^2<br />BE^2=(a/2)^2<br />=>a^2=20c^2<br />----------------------------------------<br />By Pyth. Theorem (Triangle BAE)<br />BE^2+AB^2=AE^2<br />AE^2=1.25a^2=25c^2<br />AE=5c<br />AG=AE-GE=4c<br />----------------------------------------<br />Let AM & HN meet at X and by mid-pt. theorem AX=XM=2c<br />Similarly, NX=2c<br />So by Pyth. Theorem on triangle MXN, b^2=8c^2<br />b^2/a^2=2/5<br />b/a=sqrt(2/5)<br />b=[sqrt(10)/5]*aMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68331942595189876842021-05-18T09:12:52.804-07:002021-05-18T09:12:52.804-07:00Read V as Sqrt
BF=aV5/2
Area of //gm BHFD=BF.MN/V2...Read V as Sqrt<br />BF=aV5/2<br />Area of //gm BHFD=BF.MN/V2=a^2/2<br />=>MN.aV5/2V2=a^2/2<br />=>MN=aV2/V5=aV10/5Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42236361811008527472020-10-09T23:19:29.182-07:002020-10-09T23:19:29.182-07:00Tr.s BCF & ABE are congruent SAS so AE is perp...Tr.s BCF & ABE are congruent SAS so AE is perpendicular to BM and similarly to HD<br /><br />Hence AM = a^2/ (sqrt5. a/2) = 2a/(sqrt5)<br /><br />Since HD also bisects AM, b = (AM/2). (sqrt2) = sqrt(2/5).a = a.(sqrt10)/5<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87608982322644659932020-10-07T20:07:08.856-07:002020-10-07T20:07:08.856-07:002 proofs by area method :
P the intersection of B...2 proofs by area method :<br /><br />P the intersection of BF and CG, Q that of AE and DH.<br /><br />1/ Rotating MBE clockwise around E by an angle of π, M->M’, B->C, E->E, yields a square MM’CP the same size as the central square MPNQ.<br />Doing the same with PCF around F, NDG around G and QAH around H yields a total of 5 identical squares for ABCD.<br />Therefore, a.a = 5.b.b/2 or b = a.sqrt(10)/5 QED<br /><br />2/ ABE = ABCD/4 = 5*EMB (since EMB = HQA = ABM/4)<br />So square MPNQ = 4*EMB = ABCD/5 = a*a/5, has a side equal to a*sqrt(5)/5 and its diagonal b = a*sqrt(10)/5 QEDGreghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91007222428164574072020-10-01T04:45:45.318-07:002020-10-01T04:45:45.318-07:00See the drawing
Define O intersection of AE and H...See the <a href="https://drive.google.com/file/d/1ShXR4VevkqMeUUIK35Vbjw7sFxbgqDta/view?usp=sharing" rel="nofollow"><b>drawing</b></a><br /><br />Define O intersection of AE and HD and P intersection of BF and GC<br />ABE is congruent to CBF by rotation of pi/2<br />=> x = 2 y<br />S= a^2 = 2 𝑥^2+ 3 𝑥^2 = 5 𝑥^2<br />=>𝑥^2 = a^2/5<br />b hypotenuse =>b^2= 2𝑥^2 = 2a^2/5<br />Therefore b^2 = 10 a^2/25 rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37112561384556714392020-09-30T23:04:37.063-07:002020-09-30T23:04:37.063-07:00GN=(1/2)ND=>PN=(2/5)ND (H,P,N,D).=>b=((2v2)/...GN=(1/2)ND=>PN=(2/5)ND (H,P,N,D).=>b=((2v2)/5)HDc.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9848547437513296682020-09-30T15:17:52.284-07:002020-09-30T15:17:52.284-07:00This comment has been removed by the author.rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67282634023197196572010-12-25T00:58:43.621-08:002010-12-25T00:58:43.621-08:00Joining HF, it is easy to see that
Area of paralle...Joining HF, it is easy to see that<br />Area of parallelogram BHDF <br />= sum of areas of triangles AHD & BCF <br />= (a^2)/2 <br />For the parallelogram<br />Base = HD = a (sqrt 5)/2 , <br />Height = b/(sqrt 2) &<br />Area = [a(sqrt 5)/ 2] x (b / sqrt 2)<br />So ab(sqrt 5)/2(sqrt 2) =(a^2)/2,<br />b = a(sqrt2)/(sqrt5) = a(sqrt 10)/5Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18349782073208927182010-12-21T20:03:24.174-08:002010-12-21T20:03:24.174-08:00It is so simple Anonymous
Since you said that the ...It is so simple Anonymous<br />Since you said that the inner square is (1/5) the are of the outer square which has an area of a*a<br />This means that the inner square has an area of <br />(a*a)/5<br /><br />Assuming that the left lower corner of the inner square is Q we have<br />MQ = NQ = a/sqrt(5)<br /><br />So b = sqrt(MQ^2 + NQ^2) = sqrt(a^2/5 + a^2/5)<br />= sqrt(2a^2/5).<br /><br />Multiplying both numerator and denominator within the sqrt by 5, yields a*sqrt(10)/5<br />which is the answer.Unknownhttps://www.blogger.com/profile/07363309698823587597noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14082407466101953402010-09-06T09:17:24.556-07:002010-09-06T09:17:24.556-07:00The area of the newly formed inner square is 1/5 t...The area of the newly formed inner square is 1/5 the area of the original square because:<br />1-there are 4 copies of the same triange (AMB), 3 have been rotated by 90degrees, 180 degrees and 270degrees respectively<br />2 - each of those has an area of (AM)(MB)/2<br />3 - those triangles can be arranged so that the hypotenuse (a) becomes the side of the square<br />4 - This leaves the original square with a square "hole" in the center with an area of (AM-MB)^2<br />5 - The area of four triangles is 4*[(AM)(MB)/2] or 2(AM)(MB)<br />6 - The area of the entire square is equal to the area of the small square plus the area of the four triangles, so a^2=2(AM)(MB)+(AM-MB)^2<br />7- we also know that (AM)^2 + (MB)^2 = a^2 because the triangles are right triangles<br /><br />What I can't figure out is how to get to my final answer which should be (AM-MB)^2 = 1/5a^2??<br /><br />Any ideas or help?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65335223424185209702009-06-16T22:30:26.294-07:002009-06-16T22:30:26.294-07:00Let MA and HN intersect at P.
Let NC and MF inters...Let MA and HN intersect at P.<br />Let NC and MF intersect at Q.<br />MPNQ is a square.<br />We can show that MP = AM - BM<br />So,<br />MP*MP = AM*AM + BM*BM - 2*AM*BM<br /> = a*a - 2*AM*BM<br />In triangle ABE,<br />1/[AB*AB] + 1/[BE*BE] = 1/[BM*BM]<br />So, <br />5/ [a*a] = 1/[BM*BM]<br />BM = a/ [square root of 5]<br />AM * AE = AB*AB<br />so, AM * [square root of 5]*a/2 = a*a<br /> AM = 2*a/[square root of 5]<br /><br />So,<br />MP*MP = a*a - 4*a*a/5<br /> = a*a/5<br />So,<br />MN*MN = 2*a*a/5<br /> = 10*a*a/25<br />So, <br />MN = [square root of 10]*a /5Mrudul M. Thattenoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51955640746487970392009-03-26T00:36:00.000-07:002009-03-26T00:36:00.000-07:00Let A be (0,0) then we've B;(0,a),C:(a,a) &...Let A be (0,0) then we've B;(0,a),C:(a,a) & D:(a,0). With routine analytical geometry one can determine DH:x+2y=a and CG:2x-y=a which gives us N:(3a/5,a/5). Likewise, we find M:(2a/5,4a/5).<BR/>Hence, MN^2 = b^2 = (3a/5-2a/5)^2+(a/5-4a/5)^2<BR/> = 2a^2/5 = 10a^2/5^2<BR/>Hence, b=(a/5)*(10)^(1/2)<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com