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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 877.
Triangle CEF ~ Triangle AED ~ Triangle ABFLet CE = k, then ED = 3k, AD = 4k,By Pythagorean theorem on triangle AED, (4k)^2 + (3k)^2 = 30^2 k = 6, x = 4k = 24
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.
From similar triangles we have CE=x/4 and ED=3/4.xlet θ= angle(EAD) so tan(θ)=ED/AD=4/5so cos(θ)=4/5 and x/30=4/5 => x=120/5=24
Tr. CFE///Tr.BFA so CF = x/3 & CE =x/4 and thus(x/3)²+(x/4)²=10² giving x=24