GreenlightTestPrep wrote:
In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?
Let
x = the length of AF
This means AD =
x + 5 = radius of the circle.
This is convenient, because AE is also a radius of the circle.
So, AE must have length
x + 5
At this point, we can focus on the RIGHT TRIANGLE below:
When we apply the Pythagorean Theorem we get: (
x + 5 )² +
x² = 25²
Expand to get: x² + 10x + 25 + x² = 625
Simplify: 2x² + 10x + 25 = 625
Set equal to zero: 2x² + 10x - 600 = 0
Divide both sides by 2 to get: x² + 5x - 300 = 0
Factor: (x + 20)(x - 15) = 0
So, EITHER x = -20 or x - 15
Since x cannot be negative, we know that x = 15
So, let's add this to our diagram.
Also, notice that I added some symbols to represent the 3 angles in
∆EAFAt this point we might recognize that
∆EAF and that
∆ECB are
similar trianglesHere's why:
Since EB is the diameter of the circle, we know that ∠C is 90°
Also, both triangles have ∠E in common.
If
∆EAF and
∆ECB share two angles, then the 3rd angles must also be equal.
So, the two triangles must be similar.
Let y = the length of CF
So, side EC has length
25 + yIf two triangles are similar, the ratio of their corresponding sides must be equal. We can write:
25/
40 =
20/
(25 + y)Cross multiply to get: 25(25 + y) = (40)(20)
Expand: 625 + 25y = 800
Then: 25y = 175
So y = 7
In other words, CF = 7
Answer: 7
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep