Question

PQRS is a parallelogram in which QR is produced to E such QR=RE. PE intersects SR at F. If AR(∆SFQ)=3cm square, find ar(PQRS)

Answer 1

By mid-point theorem in (PEQ)
F is the mid point of PE.
[Since R is mid point of QE (QR=RE) and SR||PQ (||gm PQRS).]
So
=>2RF=PQ.
PQ=SR (||gm)
=> 2RF=SR
Hence F is the midpoint of SR.

In (QSR) QF is the median. Hence SQF=QFR= 3cm square

In || PQRS, diagonal QS divides it into two equal triangles i.e. QPS=QSR
QSR= QSF+QFR
=> 3+3= 6 cm square.
2 ar(QSR) = ||gm PQRS (QPS=QSR)
=> PQRS= 12 cm square.

Hence the area of parallelogram PQRS is 12 cm square.

Hope this helped!