XAT 2014 — QA & DI Question 17

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

3xy = 90 ⇒ xy = 30

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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

$A(△MEV)=\frac{1}{2}\times EV\times h_1=\frac{1}{2}\times \frac{x}{4}\times \frac{y}{10}=\frac{30}{60}=0.375$

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

$A(△VFN)=\frac{1}{2}\times FV\times h_2=\frac{1}{2}\times \frac{x}{4}\times \frac{y}{14}=\frac{30}{112}\approx 0.27$

$A(□PQE)=\frac{1}{2}\times (PQ+EF)\times \frac{y}{2}$

$=\frac{1}{2}\times \left(x+\frac{x}{2}\right)\times \frac{y}{2}=\frac{3xy}{8}=\frac{90}{8}=11.25 \mathrm{sq}.\mathrm{units}$

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

= 11.25 – 0.375 + 0.27 ≈ 11.145

Hence, option (d).