SOLUTION: The figure ABCD is a rectangle, AD = 6, AB = 18, arc AB is a semicircle with diameter AB, arc CF is a semicircle with diameter CF, and EG is tangent to both semicircles. What doe

Algebra ->  Finance -> SOLUTION: The figure ABCD is a rectangle, AD = 6, AB = 18, arc AB is a semicircle with diameter AB, arc CF is a semicircle with diameter CF, and EG is tangent to both semicircles. What doe      Log On


   

Question 1149754: The figure ABCD is a rectangle, AD = 6, AB = 18, arc AB is a semicircle with
diameter AB, arc CF is a semicircle with diameter CF, and EG is tangent to both
semicircles. What does EC = ?
Image: https://imgur.com/P8UIrTR
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The drawing below is to scale.  The one given is not.


Let O be the center of the larger semicircle.
Let P be the center of the smaller semicircle.
Draw OP which will pass through the common point of tangency T and will
be perpendicular to tangent EG.
Let the radius of the small semicircle be r = CP = TP = FP
Since AB = 18, the diameter of the larger semicircle, the radius = OB = 9.
OB² + BP² = OP²
9² + (BC-CP)² = (OT+TP)²
9² + (6-r)² = (9+r)²
Solve that and get r = CP = TP = FP = 1.2
Triangles OBP and GTP are similar right triangles for they have
a common acute angle at P.  So
%28PG%29%2F%28PO%29+=+%28TP%29%2F%28PB%29
%28PG%29%2F%28OT%2BTP%29=%281.2%29%2F%28BC-CP%29
%28PG%29%2F%289%2B1.2%29=1.2%2F%286-1.2%29
Solve that and get PG=2.55
CG = CP + PG
CG = 1.2+2.55
CG = 3.75

Triangles GTP and GCE are similar right triangles for they have
a common acute angle at G.  And since right triangles OBP and GTP 
are similar, GCE and OBP are similar, so

%28EC%29%2F%28CG%29+=+%28BP%29%2F%28OB%29
%28EC%29%2F%28PG%2BCP%29=%28BC-CP%29%2F%28OB%29
%28EC%29%2F%282.55%2B1.2%29=%286-1.2%29%2F9
Solve that and get EC=2

Edwin