0 0 votes Answer the question on the basis of the information given below:In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points R and S respectively. This common tangent meets the line P and Q at O. The diameters of I and II are in the ratio $4:3.$ It is also known that the length of PO is $28$ cm.The length of SO is$8 \sqrt{3}$ cm$10 \sqrt{3}$ cm$12 \sqrt{3}$ cm$14 \sqrt{3}$ cm Quantitative Aptitude cat2004 quantitative-aptitude geometry + – go_editor 14.2k points 3.7k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
1 1 vote PR/QS =2/1.5 We know, PR/QS=PO/QO (PRO and QSO are similar triangles) or, QO=28⨉1.5/2=21 Now for a right angle triangle $QS^2 +SO^2=QO^2$ Here,SO=$\sqrt{21^2-(1.5)^2}=20.94$ Hence answer C) srestha answered Jun 24, 2016 srestha 5.2k points comment Share Follow See all 2 Comments 2 2 Comments reply Arjun 8.1k points commented Jun 26, 2016 reply Follow flag But final answer is not exactly the value in C option :O 1 1 replyShare srestha 5.2k points commented Jun 27, 2016 reply Follow flag yes 12√3 =20.78 i.e. most close to the answer So, I think that will be answer 1 1 replyShare Please log in or register to add a comment.