In a triangle ABC, points P1, Q1, R1 divide the lines AB, BC and AC respectively in the ratio 3 : 2 each. In the triangle P1 Q1 R1, the points P2, Q2 and R2 divide the sides P1Q1, Q1R1 and P1R1 in the ratio 3 : 2 each. A new triangle is constructed by joining the points on the sides of triangle P2Q2R2 that divide these sides in the ratio 3 : 2. This process of forming triangles continues infinitely. Find the ratio of the area of triangle P1Q1R1 to the sum of the areas of all the formed/constructed triangles (P2Q2R2, P3Q3R3…) whose areas are smaller than the area of the triangle P1Q1R1.

• let the original area of squareABC be P.
  P1Q1R1 area wud be 3P/2. P2Q2R2 area wud be (3/2)^2. P and so on....
  so ratio will be= {(3/2). P}/{(3/2)^2 .P+(3/2)^3 .P+ ......infinite}=P/{(3/2).P+(3/2)^2.P+..........infinite}=1/3(ans)
• 8 years agoHelpfull: Yes(1) No(5)