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Explanation:

Let the breadth be 3x and the breadth be y.

3xy = 90 ⇒ xy = 30

​​​​​​​

V is midpoint of WR. PW || EV ⇒ EV = PW/2

Similarly, FV = WQ/2

∴ EF = PQ/2 = x/2

∆MPA ∼ ∆MEV

Height of ∆MPA with respect to AP: Height (h1)of ∆MEV with respect to EV = AP : EV = x : x/4 = 4 : 1

Let height of ∆MPA = 4k and height (h1) of ∆MEV = k

∴ 4k + k = 5k = y/2

∴ k = y/10

∴ Height (h1) of ∆MEV = y/10

A(MEV)=12×EV×h1=12×x4×y10=3060=0.375

Similarly, ∆VFN ∼ ∆CRN

Height (h2) of ∆VFN with respect to VF : Height of ∆CRN with respect to CR = VF : CR = x/4 : 3x/2 = 1 : 6

Let height (h2) of ∆VFN = m and height of ∆CRN = 6m

∴ m + 6m = 7m = y/2

∴ m = y/14

∴ Height (h2) of ∆VFN = y/14

A(VFN)=12×FV×h2=12×x4×y14=301120.27

A(PQE)=12×(PQ+EF)×y2

=12×x+x2×y2=3xy8=908=11.25 sq.units

A(□PQMN) = A(□PQEF) – A(∆MEV) + A(∆VFN)

= 11.25 – 0.375 + 0.27 ≈ 11.145

Hence, option (d).

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