SOLUTION: As shown in the diagram, a circle with centre A and radius 9 is tangent to a smaller circle with centre D and radius 4. Common tangents EF and BC are drawn to the circles making po

Algebra ->  Circles -> SOLUTION: As shown in the diagram, a circle with centre A and radius 9 is tangent to a smaller circle with centre D and radius 4. Common tangents EF and BC are drawn to the circles making po      Log On


   

Question 1078647: As shown in the diagram, a circle with centre A and radius 9 is tangent to a smaller circle with centre D and radius 4. Common tangents EF and BC are drawn to the circles making points of contact at E, B and C. Determine the length of EF.
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Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!



Now we draw in line DG (in green) parallel and equal to BC.
It intersects EF at I.  DG divides radius AB = 9 into AG and BG. 
Since CD = BG = 4, AG = AB-BG = 9-4 = 5 






Now we use the Pythagorean theorem on right triangle AGD:
Hypotenuse = AD = radius AE + radius DE = 9+4 = 13

DG2 + AG2 = AD2
DG2 + 52 = 132
DG2 + 25 = 169
DG2 = 144
DG = 12

Now we have BC = DG = 12.

Next, we draw in EH parallel to AB and CD and perpendicular
to BC and G.  DG and EH are perpendicular and intersect at J 



By similar triangles EJD and AGD,

EJ%2F%28AG%29=%22DE%22%2F%22AD%22

%22EJ%22%2F5=4%2F13

13%2A%22EJ%22=20

EJ=20%2F13

HJ = CD = 4

Therefore EH = EJ+HJ = 20%2F13%2B4=20%2F13%2B52%2F13=72%2F13

Triangles IJE and EJD are similar since they are right
triangles with a common angle at D.

Triangles IJE and FHE are similar

Therefore triangles FHE and AGD are similar

%22EF%22%2F%22EH%22=%22AD%22%2F%22DG%22

%22EF%22%2F%2872%2F13%29=13%2F12

EF%2Aexpr%2813%2F72%29=13%2F12

Divide both sides by 13

EF%2F72=1%2F12

EF=72%2F12

EF=6

Edwin