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