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sujoykrdatta
A couple of questions:
1. Why ask for the GREATEST possible radius, when the exact radius can be solved for?
2. Does the question ask for the diameter?

Good questions
1) I should have just asked for the radius. That said, the answer is 80/11 either way.
I've edited the question to just ask for the radius.

2) In my original "solution" I said the length of blue dotted line was 2.5r when it's clearly 5 radii (i.e.. 5r)

I've edited both my solution AND the answer choices.
Good catch!!

Cheers and thanks,
Brent
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GMATPrepNow


The above diagram shows 3 identical circles placed inside a right triangle so that all 3 circles are tangent to line AB. What is the radius of each circle?

A) 5
B) 6
C) 7
D) 8
E) 10

Solution



    • Let the radius of each circle be r.
    • It is given that AB is tangent to all three identical circles, So, radius of each circle will be perpendicular to AB at the point of tangency.
      o So, DE is perpendicular to AB.
      o Also, \(AG = DE = r\)
         Thus, as shown in the below diagram, AEDG is a rectangle,

          • And, \(AE = GD = 5r\)
          • Also, \(∠CGD = 90\)°, and triangle CGD is a right-angled triangle.
    • Now, using Pythagoras theorem in triangle, CGD, we get:
      o \(CD^2 = CG^2 + GD^2 ⟹ CD^2 = (30-r)^2 + (5r)^2 …………Eq.(i)\)
    • Now, in the right-angled triangles DEB and DFB,
      o Side \(DE = DF = r\) and
        o hypotenuse is common, i.e. DB
        o So, EB must be equal to FB i.e. \(EB = FB = 40 – 5r\)
    • So, \(FC = BC -BF = 50 – (40 – 5r) = 10+ 5r \)
    • Now, using Pythagoras theorem in triangle in triangle, CDF,
      o \(CD^2 =DF^2 +FC^2 ⟹ CD^2 = (10+5r)^2 + (r)^2 ……………….Eq.(ii)\)
    • From Eq.(i) and (ii), we can write,
      o \((30-r)^2 + (5r)^2 = (10+5r)^2 + (r)^2\)
      o Or, \(900 + r^2 – 60r + 25r^2 = 100 + 25r^2 + 100r + r^2\)
      o Or, \(160r = 800 ⟹ r = 5\)
Thus, the correct answer is Option A.
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