Re: geometry qstn!!!..help needed?? | |
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Draw DG || RF which meets PQ at G Now FG=QG as D is the midpoint theorem using midpoint theorem: read here http://www.quantexpert.co.in/quantnotes/geometry/geometry4.php#intercept-theorem Now,FG=3/2 Again using similarity of triangles, PF/FG=4/1=> PF/(3/2)=4=> PF=6 Hence PQ=6+3=9 |
Re: geometry qstn!!!..help needed?? | |
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Hi arsh, I'm getting 9 as ans acc to the approach given by kamal sir. Please confirm the ans. |