given a rectangle ABCD , there's a point x on side AB and two points P and Q are on side BC and DE respectively, what is the ratio of area of figure PCDQ and and area of triangle XCD
i> BP=QD
ii>AQ=PC

• I am Extremely Sorry, Previously I did It Wrongly So You Can Ignore It.
  A____X____B
  | |
  Q| |D
  |_________|
  C D
  
  From S1: BP = QD & From S2: AQ = PC .
  From S1 & S2 --> PC/BP = AQ/QD = x/y(ratio)
  So Area of Triangle(XCD) / Area of Rectangle(PCDQ) = 0.5(AD*DC)/(QD*DC)
  => (0.5 AD)/QD => 0.5[(AQ+QD)/QD] => 0.5[(AQ/QD)+ 1] => 0.5[(x/y) +1 ] by. Putting the value of AQ/QD [Note:Inverse the above to get the exact Answer]
  So Answer is C) The data in both statements I and II together are necessary to answer the question.
  Note: Try To Ignore my Previous Post , This one is 100% Right Solution.
  
  
  
• 9 years agoHelpfull: Yes(22) No(1)
• Since BP=QD & AQ=PC; it is clear that if point P divides side BC in the ratio x:y then Q divides side DA in y:x. Thus, area of PCDQ=PC*CD=y*CD and that of triangle XCD=(CD*AD)/2=(CD*(x+y))/2.
  Their areas are in the ratio 2y/(x+y)
• 9 years agoHelpfull: Yes(6) No(0)
• DRAWing the figure first and let length=l and breadth=b
  area of triangle(CDX)=1/2*b*l(bcoz length will be the height and breadth will be base of the triangel )
  now take 1st condition BP=QD(suppose point P and Q are mid point)
  the area of fig PCDQ= breadth(b)*length(l/2)
  now if we move the point P towards B, then we should move Q towards D such that BP=QD so they are moving proportionally
  let P coinsides with B and Q coinsides with D forming a triangle having base b and height l,so area =1/2*b*l
  thus the area of the fig PCDQ will be (l*b)/2 just by the condition BP=QD independent of the point P and Q
  so (area of PCDQ)/(area of XCD) will be 1/1
  
• 9 years agoHelpfull: Yes(3) No(0)
• The Figure will look like a Triangle(XCD) inscribed in a Rectangle (ABCD)
  Thus to find out the PCDQ:XCD, we need to find out AQ:QD & BP:PC ,then only we can find out the areas of XCD and PCDQ,the easy way to do it if we know that Q and P are the mid points of AD & BC.Lets check that first
  (Minor Corrections P and Q lies on BC & AD)
  
  From S1: BP = QD it can't gives us the ratio
  From S2: AQ = PC it can't gives us the ratio
  
  Putting S1 and S2 together:
  AQ + QD = BP + PC [Its irrelevant actually]
  Suppose if BP = 1 & PC = 1 So , AQ = 1 & QD = 1 then ratio will be AQ:QD = 1:1 & BP:PC = 1:1
  but wait what if BP = 2 & PC = 1 [It can be possible], So AQ = 1 & QD = 2 then the ratio will be AQ:QD = 1:2 & BP:PC = 2:1.
  Again lets take BP = 7 & PC = 3 ,So AQ = 3 & QD = 7 then the ratio will be AQ:QD = 3:7 & BP:PC = 7:3.
  
  Damn it ,getting variable Ratios Thus Everything Goes Wrong ,So data is insufficient.
  Thus Answer will be d)Both statements I and II together are not sufficient to answer the question.
  
  
• 9 years agoHelpfull: Yes(2) No(0)
• well i think answer is either statement is sufficient to answer the question as beacause if statement 1 is true then 2 is automatically true and vice versa
  
  if i take the first one BP=DQ
  BC-CP=AD-AQ ( AD=BC)
  So it implies CP = AQ (AD & BC will cancel out)
  so P and Q are mid points of BC and AD
  area of PCDQ=DC*PC=1/2(DC*BC)=(1/2)area of ABCD
  AREA OF DXC=(1/2)*DC*BC=(1/2)area of ABCD
  
  so both area are equal and ratio is 1:1
• 8 years agoHelpfull: Yes(2) No(0)
• Area of PCDQ= (QD+PC)*DC/2
  since PC= AQ SO QD+PC=AD
  Hence Ar(PCDQ)= AD*DC/2
  Ar(XCD)=(1/2)*(DC)*AD
  Ar(PCDQ)/Ar(XCD)=1:1
• 9 years agoHelpfull: Yes(1) No(1)
• bhai thanks yehi sab question main bhi post karne wala tha
  
• 9 years agoHelpfull: Yes(0) No(0)
• if you remember other questions , do post , bhai
• 9 years agoHelpfull: Yes(0) No(0)
• ii>AQ=PC since both are on opposite side
  
• 9 years agoHelpfull: Yes(0) No(0)
• i remember the logarithm question in which log (base 3)N =K AND N
• 9 years agoHelpfull: Yes(0) No(0)
• 1>BP=QD by assuming P & Q are center point
• 9 years agoHelpfull: Yes(0) No(0)
• GAURAV DHIMANKAR your solution is again worng
  ! area of triangle between a rectangle is half the of the area of rectangle.
  !! and area of figure ABPQ= Area of figure QPCD
  so Area of QPCD = half of rectangle
  so Ratio of triangle and quadrilaateral =1:1
  so ANY of 1and 2 equation is enough to answer the question
• 8 years agoHelpfull: Yes(0) No(0)