The side AB of parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. Show that ar (ABCD)= ar (PBQR)
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ABCD is a parallelogram in which AC is its diagonal.
As we know that diagonals of parallelogram divides it into two triangles that are having equal areas.
∴area(ΔABC) = 1/2area ( ║gm ABCD )
so for PQBR , PQ is the diagonal.
∴area ( ΔPQB) = 1/2 (║gm PQBR)
ΔAQS and ΔAQP are on the same base.
Hence , area ΔACQ = area ΔAQP
∴areaΔACQ- ΔABQ= area ΔAQP= areaΔAQB
so ,
area ΔABC = ΔPBQ.
As we know that diagonals of parallelogram divides it into two triangles that are having equal areas.
∴area(ΔABC) = 1/2area ( ║gm ABCD )
so for PQBR , PQ is the diagonal.
∴area ( ΔPQB) = 1/2 (║gm PQBR)
ΔAQS and ΔAQP are on the same base.
Hence , area ΔACQ = area ΔAQP
∴areaΔACQ- ΔABQ= area ΔAQP= areaΔAQB
so ,
area ΔABC = ΔPBQ.
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