Question

Plz answer the first two top written lines of this paper along with the figure under it. PLZ DO IT FAST. IT IS CLASS 9 TH CBSE .

Answer 1

since ∆SPF and ∆SQF are on same base SF and between same ||s SF and PQ,
therefore, ar(SPF) = ar(SQF)
so, ∆SPF = 3 cm sq

now, in ∆SPF and FRE:
PS = RE (bcz,PS=QR=RE)
angle SPF = REF (alt. int. angles)
angle PSF = FRE (alt. int. angles)

so ∆SPF ≈ ∆FRE (by ASA cong. rule)
so ar(SPF) = ar(FRE) (bcz cong. ∆s have equal area)

so area of ∆FRE = 3 cm sq

now in ∆FEQ,
since, RE = RQ (given)
so,FR forms the median.
and a median divides a ∆ into 2 ∆s of equal area.

so area of FRQ = 3cm sq

now, adding area of ∆SQF and FRQ gives 6cm sq.

SQR forms the half of PQRS.

so area of PQRS = 6×2 = 12cm sq.
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hope this helps you.☺
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