Pl slove this

Answer :

From given information we form our diagram , As  :



Here , we know

AP  =  BP  =  12AB
And
BQ  =  CQ  =  12BC
And
AR  =  CR  =  12AC

And

APQR is a parallelogram

And from converse of mid-point theorem we know if we join mid point of two sides , that the joining line is parallel to the third line .

Here we get ABC , PBQ and RQC are similar to each other .

So , we get

Area of parallelogram APQRArea of  ABC = Area of  ABC - Area of  PBQ - Area of  RQCArea of  ABC Area of  ABC - Area of  PBQ - Area of  RQCArea of  ABC = BC2 - BQ2 - QC2BC2 Area of parallelogram APQRArea of  ABC= BC2 - BQ2 - QC2BC2 Area of parallelogram APQRArea of  ABC= BC2 - BC22 - BC22BC2 Area of parallelogram APQRArea of  ABC= BC2 - BC42 - BC42BC2 Area of parallelogram APQRArea of  ABC=   4BC2 - 2BC2 4BC2 Area of parallelogram APQRArea of  ABC=   2BC2 4BC2 Area of parallelogram APQRArea of  ABC=   2BC2 4BC2 Area of parallelogram APQRArea of  ABC=   1 2 Area of parallelogram APQR=   1 2×Area of  ABC

So, we can say that

Area of parallelogram APQR is 12 of the ABC( answer)
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