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A is a point of the side PQ of a parallelogram PQRS such that AQ=2AP. A line through Q parallel to AS intersects SR at T and PS produced at U. Then, (PS+PQ3+TQ) is equal to

A
One-third of perimeter of ΔUST
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B
Half of perimeter of ΔUST
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C
Perimeter of ΔUST
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D
Twice of perimeter of ΔUST
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Solution

The correct option is B Half of perimeter of ΔUST
Given that PQRS is a parallelogram and AQ = 2AP.
PQ=AP+AQ
=AP+2AP
=3AP
AP=PQ3
Since ST||AQ and QT||AS.
AQTS is a parallelogram
ST=AQ=2AP=2PQ3 .......(i)

Let the mid-points of US and UT be M and N, respectively. Joining M and N.
In ΔUST, by mid-point theorem,
MN||ST and MN=12ST
MN=12×AP [From (i)]
MN=AP .....(ii)
Now, in ΔUMN and ΔSPA, MN||AP and AS||UN.
MUN=PSA (Corresponding angles)
and, UMN=SPA (Corresponding angles)
Also, MN=AP [From (ii)]
ΔUMNΔSPA (AAS congruence rule)
MU=PS (CPCT) …..(iii)
and UN=SA (CPCT) …..(iv)
UN=TQ ( AQTS is a parallelogram)
From (iii), US=2MU=2PS (M is the mid-point of US)
From (iv), UT=2UN=2TQ (N is the mid-point of UT)
Now, perimeter of ΔUST=US+ST+UT
=2PS+2PQ3+2TQ
=2(PS+PQ3+TQ)
PS+PQ3+TQ=12Perimeter of ΔUST
Hence, the correct answer is option (b).

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