There are two concentric circles, between them a running track is there, Milkha singh can run 24m straight starting from one point on the outer circle to the other point on the outer circle touching only once the inner circle. What is the amount of synthetic material required to build the track. (Asked on 8th June)
A) 144 pi B) 162 pi C) 192 pi

• draw a st. line from a point on outer circle touching inner circle & passing through other end of outer circle. it will be a tangent of length (24 cm) to smaller circle.
  draw a perpendicular to this tangent, it will divide tangent in two equal parts.
  
  if 'r' be radius of inner circle & 'R' of outer circle then
  R^2=(12^2+r^2)
  
  amount of synthetic material required = pi(R^2-r^2)=Pi*[(12^2+r^2)-r^2] = 144pi
• 10 years agoHelpfull: Yes(71) No(4)
• A) 144 pi
  
  Milkha singh can run along an eq. triangle of side 24 cm ,for which inner circle will be incircle of triangle.
  if r be radius of inner circle then
  area of triangle = r*s where,s =semi perimeter of triangle
  => √3/4 * 24*24 = r*(24+24+24)/2 => r = 4√3 cm
  if R be radius of outer circle then, R =√(12^2+r^2) = √192 cm
  
  amount of synthetic material required = pi*(R^2-r^2)=pi*(192-48)= 144pi
• 10 years agoHelpfull: Yes(19) No(2)
• draw a straight line from a point on outer circle touching inner circle & passing through other end of outer circle...draw the diagram properly...a triangle will form if we join the the two points touching the outer circle and the center of those two circles...the newly drawn two sides of the triangle are two radius of the outer circle...let radius of outer circle is r....now from the center draw a straight line to that point of the inner circle where the line drawn of 24 mt is touching the edge of the inner circle...now this newly drawn line is the radius of inner circle...so, the tangent of 24 mt will be divided into two equal parts of 12 mt...consider any of the small triangles with base of 12 mt, perpendicular of x ( the radius of inner circle) and radius of outer circle r....
  
  so, tan 45=(x/12), x=12..sin 45=(x/r)...r=12√2...
  required ans= the area of outer circle - the area of inner circle
  =(pi r^2 -pi x^2)
  =pi(12√2^2 - 12^2)
  =pi(288 -144) = pi 144 (ans)
• 10 years agoHelpfull: Yes(17) No(3)
• 144 pi is correct ...
  let R is big redi and r is small redi..
  R= sqrt(r^2 + 12^2)
  
  required material= pi(R^2 - r^2) put the value of R then = pi*144
  
  logic - redi is normal to its tangent. so draw the diagram u will see.
• 10 years agoHelpfull: Yes(2) No(0)
• draw concentric circle, when milkha singh runs in straight line he forms a tangent with the inner circle
  apply pythagoras theorm R^2 = 144 + r^2
  area needed = 3.14(R^2-r^2) =3.14*144 = 144 pi
• 10 years agoHelpfull: Yes(1) No(1)
• R^2=r^2+(24/2)^2
  R^2=r^2=144
  area of track=R^2pi-r^2pi
  =144pi
• 10 years agoHelpfull: Yes(1) No(0)
• When we draw the required fiq we will find that there are concentric circles with a chord of larger circle which is tangent of smaller circle( as it is given that Mikha singh touches only once the inner circle)
  And it is one of the property of two concentric circles that the chord of the larger circle that is tangent to the smaller circle is bisected at the point of contact.
  now from figure we will get a right angle triangle with one of the side 12 cm( with angle 90).
  Now if r be the radius of inner circle and R be the radius of outer circle, we will get
  R^2=(12^2+r^2)
  so, the required amount of synthetic material= pi*(R^2-r^2)
  =Pi*[(12^2+r^2)-r^2]
  = 144pi
• 9 years agoHelpfull: Yes(1) No(0)
• when we draw the two concentric circles , one is having greater radii i.e R and smaller one with r radii. acc to second information given that milkha singh can rum 24m straight from one point of outer circle to another point of outer circle and touches only one time with inner circle i.e tangent to the inner circle...if we join the three tangents which touches inner circle forms a equilateral triangle. name as ABC ,if we draw a median from A to the third side BC it bisect BC at a point say R.. if notice that median AR is equal to the sum of R+r ,as we if three medians draw from A B C meet at centroid which divides medians in 2:1 , say centroid be P ,AP= R ,PR=r ,in equilateral triangle AP= side/root3 and PR= side/2*sq root3 put side as 24 ,AP=8sq root3 and PR=4sqroot3 ,required area= pi[R^2-r^2]=pi[8sqroot3^2 - 4sqroot3^2]= pi[64*3-14*3]=pi144 is the ans.
• 10 years agoHelpfull: Yes(0) No(1)