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

4 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

    ReplyDelete
  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.

    ReplyDelete
  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

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

    ReplyDelete