Given 3 lines in the plane such that the points of intersection form a triangle with sides of length 20, 20 and 30, the number of points equidistant from all the 3 lines is
a) 4
b) 3
c) 1
d) 0

• only one point i.e. In center of the triangle
  
• 14 years agoHelpfull: Yes(12) No(6)
• 4 points equidistant
  one is center point of the triangle
  all the other is out side of each vertices
• 14 years agoHelpfull: Yes(9) No(2)
• plz explain friends. i m confused
• 14 years agoHelpfull: Yes(5) No(2)
• 1
• 14 years agoHelpfull: Yes(3) No(3)
• 1
• 14 years agoHelpfull: Yes(2) No(2)
• 1 point
• 14 years agoHelpfull: Yes(2) No(3)
• 1
• 14 years agoHelpfull: Yes(2) No(3)
• 0 will be the answer because only equilateral triangle will have equidistant to its center. Here (20,20,30) sides doesn't form equilateral triangle
• 14 years agoHelpfull: Yes(1) No(6)
• There are 4 such points. One point is the incenter of the triangle.
  3 are the excenters with respect to each angle of the triangle.
• 9 years agoHelpfull: Yes(0) No(0)