PQRS is a parallelogram, QR is produced to E so that QR=RE and PE intersects RS in F. If ar(QSF)=4cm square find ar(PQRS)

PQRS is a parallelogram, QR is produced to E so that QR=RE and PE intersects RS in F. If ar(QSF)=4cm square
, find ar(PQRS)

3

Answers


  1. 0 Votes

    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).

  2. 0 Votes

    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.]

  3. 0 Votes

    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.

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