Math Problem - Help needed to solve

[priyasaibaba]

Can anyone help me solve this problem?

As the figure above shows, two circles lie on a line and tangent to each other. If the radius of two circles are 2r and r, respectively, in terms of r, AB=?
(A) 5r/2
(B) 8r/3
(C) 2*2^1/2*r
(D) 3r
(E) 2*3^1/2*r

Pls find the figure in the attached doc.

Thanks for your help.

[eaakbari]

Do attach the figure, you will not be able to paste it. I cannot understand the figure solely through your explanation

[priyasaibaba]

Hi eaakbari, I have attached the figure. Pls let me know, if u find the answer.

Thanks,
Priya

[priyasaibaba]

D is not the answer. C is the answer. Can someone explain the steps to get the answer?

[eaakbari]

Answer is C. wait let me explain

[eaakbari]

CD is a line joining the 2 centers
ED is a line parallel and equal to AB which meets line AC at point E

Now CD = 2r + r = 3r
CE = CA-EA = 2r - r = r

Hence Using pythogoras

ED^2 = 9r^2 + r^2
ED = AB = ROOT(8r^2)
= 2root2 * r

HTH

[akhpad]

Correct, C is the answer

[priyasaibaba]

Wow, thanks very much. Appreciate your help.

[pkw209]

good problem. Haven't come across one like that yet.

[pkw209]

...until now

[Stuart@KaplanGMAT]

CD is a line joining the 2 centers
ED is a line parallel and equal to AB which meets line AC at point E

Now CD = 2r + r = 3r
CE = CA-EA = 2r - r = r

Hence Using pythogoras

ED^2 = 9r^2 + r^2
ED = AB = ROOT(8r^2)
= 2root2 * r

HTH

Great job!

You ended up with the correct answer, just want to correct one small typo so no one gets confused along the way:

ED^2 = 9r^2 + r^2

9(r^2) is actually the hypotenuse, not ED^2, so that should be:

ED^2 = 9(r^2) - r^2

You did the math as though it were a subtraction sign, so all is good after that point!