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.
Tuesday, May 21, 2013
Problem 877: Square, Right Triangle, Similarity, Metric Relations, Transversal
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Triangle CEF ~ Triangle AED ~ Triangle ABF
ReplyDeleteLet 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.
ReplyDeleteFrom similar triangles we have CE=x/4 and ED=3/4.x
ReplyDeletelet θ= angle(EAD) so tan(θ)=ED/AD=4/5
so 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
ReplyDelete(x/3)²+(x/4)²=10² giving x=24
Using only pythagoras,
ReplyDeleteAD=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
∆CFE ~ ∆DAE
ReplyDeleteCF/DA=EF/AE=1/3
CF=x/3
CE=√(100-x^2/9),DE=√(900-x^2 )
CE+DE=CD=x
√(100-x^2/9)+√(900-x^2 )=x
(4/3)* √(900-x^2 )=x
900-x^2=(9x^2)/16
x^2=576
x=24
Easily from similar triangles
ReplyDeleteCF = AD / 3 = x/3
CE = AB / 4 = x/4
Using Pythagoras on Triangle CEF,
(x/4)^2 + (x/3)^2 = 10^2
Solving for x
x = 24
Sumith Peiris
Moratuwa
Sri Lanka