.a triangle ABC is given, a line DE is parallel to base side and that cuts the triangle.. the ratio of area of triangle to the area of trapezium .given DE/BC=3/5..

• Area(debc)/area(abc) = (area(ABC)-area(ADE))/area(ABC) = 1-area(ADE)/area(ABC)=1-3/5 = 2/5 (since ratio of areas is equal to ratio of parllel bases).
  Now area(ABC)/Area(trapezium) = 5/2.
• 12 years agoHelpfull: Yes(16) No(17)
• ANS : 9/16
  The line DE divides the triangle ABC into 2 parts - triangle ADE and trapezium DBCE.The question is to find the ratio of area(ADE) to area(DBCE).
  
  Now triangle ADE is similar to triangle ABC.
  => DE/BC = AD/AB = AE/AC ....(1)
  Drawing a perpendicular from A to BC meeting BC at G and meeting DE at F.
  Now AF and AG are the heights of the triangles ADE and ABC resp.
  tri(ADF)~tri(ABG) => AD/AB = AF/AG ....(2)
  Area of a triangle = 1/2(base*height).
  Area of tri(ADE) = 1/2(DE*AF)
  Area of tri(ABC) = 1/2(BC*AG)
  Area of trapezium(DBCE) = area(ABC)-area(ADE) = 1/2(BC*AG-DE*AF)
  Ratio - area(ADE)/area(DBCE) = (DE*AF)/(BC*AG-DE*AF) = 1/[(BC*AG-DE*AF)/(DE*AF)]
  (rearranging) = 1/[(BC/DE)*(AG/AF)-1]
  Using (!) and (2),(BC/DE)*(AG/AF) = (BC/DE)^2
  (using given DE/BC=3/5) => BC/DE=5/3 => (BC/DE)^2 = 25/9
  The required ratio = 1/(25/9-1) =1/(16/9) = 9/16
• 11 years agoHelpfull: Yes(14) No(0)
• ANS : 9/16
  The line DE divides the triangle ABC into 2 parts - triangle ADE and trapezium DBCE.The question is to find the ratio of area(ADE) to area(DBCE).
  
  Now triangle ADE is similar to triangle ABC.
  [since
• 12 years agoHelpfull: Yes(4) No(2)
• 9/16 since the line divides the riangle in two parts
• 8 years agoHelpfull: Yes(0) No(0)