Question

# 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 ΔUSTGiven 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)] ∴ΔUMN≅ΔSPA (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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