## 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.

#### 5 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

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.

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

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

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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