Question

Q 29 please anybody solve​

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

$SQF=QFR= \: 3 {cm}^{2}$

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)

$=> PQRS= 12 {cm}^{2}$

$Hence \: the \: area \: of \: parallelogram \: PQRS is 12 {cm}^{2}$