Since, RF || PQ and R mid point of QE (?)
By converse of mid-point theorem, F will be mid-point of SR, ie, SF = RF

ar(triangle QSF) = ar(triangle RFE)= 4 cm^2, because both triangles have equal base, (SF and RF) and lies between equidistant parallels (draw a line EG from E parallel to PQ, complete the parallelogram REGS, then PQRS = RSEG).

given: PQRS is a parallelogram and QR = RE
area(QSF) = 4 sq cm
in the triangles PSF and ERF;
PS = RE [since PS = QR]
∠PSF = ∠FRE [alternate interior angles]
∠SPF = ∠REF [alternate interior angles , since SP || RQ]
therefore by ASA congruency, triangles are congruent.
and by cpct SF = FR
area(QSF) = area(SPF) [triangles between same pair of parallel and with same base are equal in areas]
area(QSF) = 1/2*SF*distance between parallel lines PQ and RS = 4
SF*distance between parallel lines PQ and RS = 4*2 = 8 sq cm…………(1)
area(PQRS) = SR * distance between parallel lines PQ and RS.
= 2* SF *distance between parallel lines PQ and RS.
= 2*8
= 16 sq cm
hope this helps you.

3

## Answers

Since, RF || PQ and R mid point of QE (?)

By converse of mid-point theorem, F will be mid-point of SR, ie, SF = RF

ar(triangle QSF) = ar(triangle RFE)= 4 cm^2, because both triangles have equal base, (SF and RF) and lies between equidistant parallels (draw a line EG from E parallel to PQ, complete the parallelogram REGS, then PQRS = RSEG).

Since, RF || PQ and R mid point of QE (?)

By converse of mid-point theorem, F will be mid-point of SR, ie, SF = RF

ar(triangle QSF) = ar(triangle QFR)= 4 cm^2, because both triangles have equal base, (SF and RF) and lies between same parallels PQ and RS.

So, ar(triangle QSR)=ar(triangle QSF) + ar(triangle QFR)= 4+4=8,

and ar(PQRS) = 2 ar(triangle QSR) = 2*8= 16 square cm

[diagonal of a parallelogram (here QS) divides it in 2 congruent traingless.]

given: PQRS is a parallelogram and QR = RE

area(QSF) = 4 sq cm

in the triangles PSF and ERF;

PS = RE [since PS = QR]

∠PSF = ∠FRE [alternate interior angles]

∠SPF = ∠REF [alternate interior angles , since SP || RQ]

therefore by ASA congruency, triangles are congruent.

and by cpct SF = FR

area(QSF) = area(SPF) [triangles between same pair of parallel and with same base are equal in areas]

area(QSF) = 1/2*SF*distance between parallel lines PQ and RS = 4

SF*distance between parallel lines PQ and RS = 4*2 = 8 sq cm…………(1)

area(PQRS) = SR * distance between parallel lines PQ and RS.

= 2* SF *distance between parallel lines PQ and RS.

= 2*8

= 16 sq cm

hope this helps you.