PR is a tangent to a circle at point P.Q is another point on the circle such that PQ is the diameter and RQ cuts the circle at point M. If the radius of the circle is 4 units and PR=6 units then find the ratio of the perimeter of triangle PMR to the triangle PQR
a)11/20
b)3/5
c)13/20
d)18/25

• The ans is (b) 3/5
  
  Sol:
  I hope the geometry of this questn is clr.
  
  Nw since PR is the tangent,Hence angle QPR is 90.
  =>Triangle PQR is rit-angled triangle.
  =>hyp QR=10. (Pythagoras Theorem)........(1)
  
  Also PM is perpendicular to QR (Since triangle PQR is made inside circle with diameter as hyp
  hence PMQ is rit-angled...itz a property of circle)
  =>three triangles PQR,PMQ & PMR are similar to each other.
  (For similarity of such type of triangles u can refer any 10th std book).
  
  NOw in that pic consider the triangles PQR & PMR.
  PQR~PMR
  hence PR/RM=QR/PR.
  =>PR^2=RM*QR
  putting the values & from(1)
  =>RM=18/5....(2)
  
  Also
  PQR~PQM
  =>PQ/QM=QR/PQ
  =>PQ^2=QM*QR
  putting the values & from(1)
  =>QM=32/5.....(3)
  
  Also,
  PQM~PMR
  =>PM/QM=RM/PM
  =>PM^2=QM*RM
  from(2) & (3)...
  PM=24/5......(4)
  
  Hence peimeter of triangle PMR=PM+RM+PR
  Putting the obtained values perimeter=72/5
  
  Also perimeter of triangle PQR=PQ+QR+PR=24.
  
  Hence th ratio PMR/PQR=72/(5*24)=3/5
  
  :-))
• 12 years agoHelpfull: Yes(86) No(9)
• mr ashu dont be oversmart
  
• 10 years agoHelpfull: Yes(14) No(3)
• b)3/5 is the ratio of the perimeter of triangle PMR to the triangle PQR.
• 12 years agoHelpfull: Yes(8) No(12)
• Answer
  
  
  Since PR is tangent, angle RPQ = 90°. Hence QPR is a right angled triangle.
  
  Given that PR = 6 units ----(1)
  PQ = Diameter of the circle = 2×4 = 8 units ----(2)
  
  Using Pythagorean theorem, QR2 = PR2 + PQ2 = 62+82 = 100
  => QR = 10 units ----(3)
  
  Angle inscribed in a semicircle is always 90° (Thales' theorem)
  => angle PMQ is 90°
  
  sin PRM = sin PRQ
  PM/PR=PQ/RQ
  PM/6=8/10
  PM = 48/10 = 4.8 units ----(3)
  
  tan PRM = tan PRQ
  PM/RM=PQ/PR
  4.8/RM=86
  RM = 3.6 units ----(4)
  
  Perimeter of triangle PMR = PM + RM + PR = 4.8 + 3.6 + 6 = 14.4 units
  Perimeter of triangle PQR = PQ + QR + PR = 8 + 10 + 6 = 24 units
  
  Perimeter of triangle PMRPerimeter of triangle PQR=14.4/24=3/5
  
• 9 years agoHelpfull: Yes(7) No(0)
• explian the ans plz
• 12 years agoHelpfull: Yes(1) No(6)
• the ans is (c)
  13/20
• 11 years agoHelpfull: Yes(0) No(7)
• Since PR is tangent, angle RPQ = 90°. Hence QPR is a right angled triangle.
  
  Given that PR = 6 units ----(1)
  PQ = Diameter of the circle = 2×4 = 8 units ----(2)
  
  Using Pythagorean theorem, QR2 = PR2 + PQ2 = 62+82 = 100
  => QR = 10 units ----(3)
  
  Angle inscribed in a semicircle is always 90° (Thales' theorem)
  => angle PMQ is 90°
  
  sin PRM = sin PRQ
  PMPR=PQRQPM6=810
  PM = 4810 = 4.8 units ----(3)
  
  tan PRM = tan PRQ
  PMRM=PQPR4.8RM=86
  RM = 3.6 units ----(4)
  
  Perimeter of triangle PMR = PM + RM + PR = 4.8 + 3.6 + 6 = 14.4 units
  Perimeter of triangle PQR = PQ + QR + PR = 8 + 10 + 6 = 24 units
  
  Perimeter of triangle PMR Perimeter of triangle PQR=14.424=144/240=3/5
• 9 years agoHelpfull: Yes(0) No(0)
• 
  
  Since PR is tangent, angle RPQ = 90°. Hence QPR is a right angled triangle.
  
  Given that PR = 6 units ----(1)
  PQ = Diameter of the circle = 2×4 = 8 units ----(2)
  
  Using Pythagorean theorem, QR2 = PR2 + PQ2 = 62+82 = 100
  => QR = 10 units ----(3)
  
  Angle inscribed in a semicircle is always 90° (Thales' theorem)
  => angle PMQ is 90°
  
  sin PRM = sin PRQ
  PM
  PR
  =
  PQ
  RQ
  PM
  6
  =
  8
  10
  PMPR=PQRQPM6=810
  PM =
  48
  10
  4810 = 4.8 units ----(3)
  
  tan PRM = tan PRQ
  PM
  RM
  =
  PQ
  PR
  4.8
  RM
  =
  8
  6
  PMRM=PQPR4.8RM=86
  RM = 3.6 units ----(4)
  
  Perimeter of triangle PMR = PM + RM + PR = 4.8 + 3.6 + 6 = 14.4 units
  Perimeter of triangle PQR = PQ + QR + PR = 8 + 10 + 6 = 24 units
  
  Perimeter of triangle PMR
  Perimeter of triangle PQR
  =
  14.4
  24
  =
  144
  240
  =
  3
  5
  Perimeter of triangle PMRPerimeter of triangle PQR=14.424=144240=35
  
• 8 years agoHelpfull: Yes(0) No(0)
• PQ=8(diameter of the circle)
  PR=6
  so, QR=10 (pythagorean theorem)
  
  angle QMP=90 degree (angle subtanding diameter of a circle)
  angle QMP=angle RMP=90 degree
  
  assume QM=a and MP=b
  MR=10-a
  
  we get equations, a^2 + b^2=8^2......(i)
  and, b^2 + (10-a)^2=6^2......(ii)
  
  solving these two equations, we get, a=6.4 and b=4.8
  
  so, perimeter(triangle PMR)/ perimeter(triangle PQR)=(b+10-a+6)/(8+10+6)=14.4/20=18/25......option (d)
• 7 years agoHelpfull: Yes(0) No(0)
• Ans:13/20
  PQ-8
  PR-6
  QR-10 by pythagores theorm
  let PM be k
  PQR right 3gle
  PM- x
  MR- 10-x
  PMR also right 3gle as formed by diameter
  64=k^2+x^2
  36=(10-x)^2+k^2
  
  solving for x gives
  x=(64/10)
  then k=(48/10)
  perimeter of PMR=6+(48/10)+(48/10)
  perimetr of PQR =6+8+10
  ratio
  13/20
• 5 years agoHelpfull: Yes(0) No(3)