Question

In the figure, ABCD is a parallelogram and CP is parallel to DB. Prove that: Area of OBPC = `(3)/(4)"area of ABCD"`

Answer 1

Since the diagonals of a parallelogram divide it into four triangles of equal area
Therefore, area of ΔAOD = area ΔBOC = area ΔABO = area ΔCDO.

⇒ area ΔBOC = `(1)/(4)`area(||gm ABCD)  ..........(i)

In ||gm ABCD, BD is the diagonal
Therefore, area(ΔABD) = area(ΔBCD)

⇒ area (ΔBCD) = `(1)/(2)`area(||gm ABCD)  .......(ii)

In ||gm BPCD, BC is the diagonal
Therefore, area(ΔBCD) = area(ΔBPC) .....(iii)
From (iii) and (ii)

⇒ area (ΔBPC) = `(1)/(2)`area(||gm ABCD)  .......(iv)

adding (i) and (iv)

area(ΔBPC) + areaΔBOC = `(1)/(2)"area(||gm ABCD)" + (1)/(4)"area(||gm ABCD)"`

Area of OBPC = `(3)/(4)"area of ABCD"`.