tag:blogger.com,1999:blog-6933544261975483399.post747294886981707809..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 877: Square, Right Triangle, Similarity, Metric Relations, TransversalAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-88619890498922764892024-03-19T00:02:30.728-07:002024-03-19T00:02:30.728-07:00Easily from similar triangles
CF = AD / 3 = x/3
C...Easily from similar triangles<br /><br />CF = AD / 3 = x/3<br />CE = AB / 4 = x/4<br /><br />Using Pythagoras on Triangle CEF,<br /><br />(x/4)^2 + (x/3)^2 = 10^2<br /><br />Solving for x<br /><br />x = 24<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30265973444283054962024-03-17T00:52:02.052-07:002024-03-17T00:52:02.052-07:00∆CFE ~ ∆DAE
CF/DA=EF/AE=1/3
CF=x/3
CE=√(100-x^2/9)...∆CFE ~ ∆DAE<br />CF/DA=EF/AE=1/3<br />CF=x/3<br />CE=√(100-x^2/9),DE=√(900-x^2 )<br />CE+DE=CD=x<br />√(100-x^2/9)+√(900-x^2 )=x<br />(4/3)* √(900-x^2 )=x<br />900-x^2=(9x^2)/16<br />x^2=576<br />x=24Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74697031235625274382018-02-14T14:18:29.529-08:002018-02-14T14:18:29.529-08:00Using only pythagoras,
AD=x=CD=BC,ED^2=900-x^2, CE...Using only pythagoras,<br />AD=x=CD=BC,ED^2=900-x^2, CE=x-root(900-x^2), CF^2=100-(x-root(900-x^2))^2, CF=root(100-(x-root(900-x^2))^2)), finally x^2+(x+root(100-(x-root(900-x^2))^2))^2=40^2, x=24<br />Anonymoushttps://www.blogger.com/profile/11988214253974032468noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7747745584931725942013-05-21T20:12:15.327-07:002013-05-21T20:12:15.327-07:00Tr. CFE///Tr.BFA so CF = x/3 & CE =x/4 and thu...Tr. CFE///Tr.BFA so CF = x/3 & CE =x/4 and thus<br />(x/3)²+(x/4)²=10² giving x=24Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85208206980938518412013-05-21T19:49:23.704-07:002013-05-21T19:49:23.704-07:00From similar triangles we have CE=x/4 and ED=3/4.x...From similar triangles we have CE=x/4 and ED=3/4.x<br />let θ= angle(EAD) so tan(θ)=ED/AD=4/5<br />so cos(θ)=4/5 <br />and x/30=4/5 => x=120/5=24Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16305544966156738582013-05-21T19:34:36.210-07:002013-05-21T19:34:36.210-07:00Triangle ABF is a 3-4-5 triangle. Thus, since the ...Triangle ABF is a 3-4-5 triangle. Thus, since the "5" side (AF) is 40 in length, the "3" side (AB) is 24. It's a simple ratio.Tom Bombadilhttps://www.blogger.com/profile/17852647625795740222noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83927061754880243512013-05-21T19:24:40.358-07:002013-05-21T19:24:40.358-07:00Triangle CEF ~ Triangle AED ~ Triangle ABF
Let CE ...Triangle CEF ~ Triangle AED ~ Triangle ABF<br />Let CE = k, then ED = 3k, AD = 4k,<br />By Pythagorean theorem on triangle AED, <br />(4k)^2 + (3k)^2 = 30^2 <br />k = 6, <br />x = 4k = 24<br />nakahttps://www.blogger.com/profile/11277356476170372732noreply@blogger.com