Q. ABCD is a square. P is the mid point of AB. the line passing through A and perpendicular to DP intersects the diagonal at Q and BC at R, If AB =2 then PR=?

• Let the line passing thru A meet DP in E.
  
  So In triangle DAP
  
  DP= Sqroot(5)
  
  AE=2/Sqroot(5)
  
  triangle AEP is similar to triangle ABR
  
  So AR= sqroot(5)
  
  =>RB=1
  
  => PR=Sqroot(2)
• 10 years agoHelpfull: Yes(11) No(17)
• let E is point of intersection of line AR.
  IN TRIANGLE ADP;
  WE GET,
  OP=SQRT(5).
  now in similar triangles AEP AND ADP (By angle p in both triangles is common and one side AP is also common)
  therefore
  AP/DP=EP/AP
  AP^2= EP*DP
  1=EP*SQRT(5)
  SO, EP=1/SQRT(5)
  AGAIN IN SIMILAR TRIANGLES APE AND PER(BY ONE SIDE EP IS COMMON)
  AP/PE=PE/PR
  AP*PR=PE^2
  1*PR={1/SQRT(5)}^2
  HENCE, PR=1/5.
  WHICH IS REQUIRED VALUE OF PR.
• 10 years agoHelpfull: Yes(4) No(4)
• Ans : square root of 2. if u draw the diagram it will give u an triangle where u can apply pythagoras theorem..
• 10 years agoHelpfull: Yes(1) No(2)
• Plesse give options
• 10 years agoHelpfull: Yes(0) No(3)
• underroot(29) / 5.
• 7 years agoHelpfull: Yes(0) No(0)