Tuesday, May 21, 2013

Problem 877: Square, Right Triangle, Similarity, Metric Relations, Transversal

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.

Online Geometry Problem 877: Square, Right Triangle, Similarity, Metric Relations, Transversal

7 comments:

  1. Triangle CEF ~ Triangle AED ~ Triangle ABF
    Let CE = k, then ED = 3k, AD = 4k,
    By Pythagorean theorem on triangle AED,
    (4k)^2 + (3k)^2 = 30^2
    k = 6,
    x = 4k = 24

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  2. 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.

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  3. From similar triangles we have CE=x/4 and ED=3/4.x
    let θ= angle(EAD) so tan(θ)=ED/AD=4/5
    so cos(θ)=4/5
    and x/30=4/5 => x=120/5=24

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  4. Tr. CFE///Tr.BFA so CF = x/3 & CE =x/4 and thus
    (x/3)²+(x/4)²=10² giving x=24

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  5. Using only pythagoras,
    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

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  6. ∆CFE ~ ∆DAE
    CF/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

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  7. Easily from similar triangles

    CF = 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

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