There is a trapezium ABCD such that the sides AB and CD are parallel to each other. The diagonals AC and BD intersect at O. The area of the triangle COD is 10 square cm and the area of the triangle AOB is 40 square cm. Find the area of the trapezium.P.s. Please post the approach as well.

1) 90
2) 100
3) 250
4) 120

• In the givem diagram draw perpendiculars from point O to CD and AB at M and N respectively.
  Now area of triangle COD = (1/2)(CD)(OM)
  = 10
  Similarly area of triangle AOB = (1/2)(AB)(ON)
  = 40
  Now AB:DC = 4:1
  Since triangle DOC and triangle AOB are similar triangles.
  Therefore ON :ON = 1:2
  So the area of trapezium = 0.5 (AB + CD)(OM + ON)
  = 90 sq.cm
• 11 years agoHelpfull: Yes(0) No(3)