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, $(PS+\frac{PQ}{3}+TQ)$ is equal to

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

Answer 1

The correct option is B Half of perimeter of ΔUST

Given that PQRS is a parallelogram and AQ = 2AP.
$\therefore PQ=AP+AQ$
$=AP+2AP$
$=3AP$
$\Rightarrow AP=\frac{PQ}{3}$
Since $ST||AQ$ and $QT||AS$.
$\therefore AQTS$ is a parallelogram
$\Rightarrow ST=AQ=2AP=\frac{2PQ}{3}$ .......(i)

Let the mid-points of US and UT be M and N, respectively. Joining M and N.
$\therefore$ In $\Delta UST$, by mid-point theorem,
$\therefore MN||ST$ and $MN=\frac{1}{2}ST$
$\Rightarrow MN=\frac{1}{2}\times AP$ [From (i)]
$\Rightarrow MN=AP$ .....(ii)
Now, in $\Delta UMN$ and $\Delta SPA$, $MN||AP$ and $AS||UN$.
$\Rightarrow \angle MUN=\angle PSA$ (Corresponding angles)
and, $\angle UMN=\angle SPA$ (Corresponding angles)
Also, $MN=AP$ [From (ii)]
$\therefore \Delta UMN\cong \Delta SPA$ (AAS congruence rule)
$\Rightarrow MU=PS$ (CPCT) …..(iii)
and $UN=SA$ (CPCT) …..(iv)
$\Rightarrow UN=TQ$ ($\because$ 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 $\Delta UST=US+ST+UT$
$=2PS+\frac{2PQ}{3}+2TQ$
$=2(PS+\frac{PQ}{3}+TQ)$
$\Rightarrow PS+\frac{PQ}{3}+TQ=\frac{1}{2}\text{Perimeter of}\Delta UST$
Hence, the correct answer is option (b).