Xavier Aptitude Test 2017 Paper

Question : 62 of 97

Marks: +1, -0

ABCD is a rectangle. P, Q and R are the midpoint of BC, CD and DA. The point S lies on the line QR in such a way that SR: QS = 1:3. The ratio of the triangle APS and rectangle ABCD is

36/128
39/128
44/128
48/128
64/128

Validate

Solution:

(a)

Area of A APS = Area of rectangle ABCD - Area Δ APB
-Area ΔQPC - Area ΔDQR- Area Δ ARS - Area Δ PQS
Let’s assume the dimensions of rectangle ABCD as length, AB =32 and width BC =24
Thus, Area

\Δ{APS}=32 \× 24-\ {1}/{2} \×32 \× 12-\ {1}/{2} \× 16 \× 12-

\ {1}/{4}

Area

Δ{ARQ}-\ {3}/{4}

Area

\Δ{PRQ}

...... (1)

Now, Area ΔARQ =Area of rectangle ADQE - Area of Δ DRQ-AreaΔAQS

=16 ×24-\ {1}/{2} × 16 \×12-\ {1}/{2} × 24 × 16

=384-96-192=96

And Area

\Δ{PRQ}=

Area of rectangle CDRP-Area

\Δ{DRQ}-

Area

\Δ{PQC}=12 \×32-\ {1}/{2} \×12 × 16-\ {1}/{2} \×12 \×16=

384-96-96=192

Thus, from equation (i), we have
Area

\Δ{APS}=768-192-96-96- {3}/{4} \×192- {1}/{4} × 96

=384

-144-24=216

Thus required ratio

=\ {\ { Area }\ Δ\ {APS}}/{\ { Area }\ Δ\ {ABCD}}

=\ {216}/{32 \× 24}= {216}/{768}= {36}/{128}

Hence option

(\ {A})

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