in Quantitative Aptitude retagged by
341 views
1 vote
1 vote

Two circles, each of radius $4$ cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is

  1. $\sqrt{2}$
  2. $\frac{\pi }{3}$
  3. $\frac{1}{\sqrt{2}}$
  4. $1$
in Quantitative Aptitude retagged by
13.4k points
341 views

1 Answer

1 vote
1 vote

Let $r$ be the radius of the third circle. Then,


From the above diagram, we get

$AE= GC + CD$

$\Rightarrow 4 = r + CD $

$ \Rightarrow CD = ( 4-r )$

In $ \triangle ADC,$



$ \angle D = 90^{\circ},$ using Pythagoras theorem.

$ \boxed{(AC)^{2} = (CD)^{2} + (AD)^{2}} $

$(4+r)^{2} = (4-r)^{2} + (4)^{2}$

$ \Rightarrow 16 + r^{2} +  8r = 16 + r^{2} – 8r + 16$

$ \Rightarrow 8r = -8r + 16 $

$\Rightarrow 8r + 8r = 16$

$ \Rightarrow 16r = 16$

$ \Rightarrow r = 1$

$ \therefore $ The radius of the third circle is $1\;\text{cm}.$

Correct Answer: D

edited by
10.3k points
Answer:

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true