# Difficult Geometry & Mensuration Solved QuestionAptitude Discussion

 Q. Through $T$, the mid-point of the side $QR$ of a  $\triangle PQR$ , a straight line is drawn to meet $PQ$ produced to $S$ and $PR$ at $U$, so that $PU = PS$. If length of $UR = 2$ units then the length of $QS$ is
 ✖ A. $2\sqrt{2}$ units ✖ B. $\sqrt{2}$ units ✔ C. $2$units ✖ D. cannot be determined

Solution:
Option(C) is correct

We have $QT = TR$ and $PU = PS. UR = 2$ units

We draw $RV \parallel PS$ that meets $SU$ extended at $V.$

In  $\triangle QST$  and  $\triangle TVR$

$\angle QTS =\angle VTR$ [Opposite angles]

$\angle QST =\angle TVR$ [Alternate angles as $PS \parallel VR$]

$QT+TR$

$\because \triangle QST$ and $\triangle TVR$ are congruent.

$\therefore QS = VR$ -------- $(i)$

Now

$\angle QST = \angle PUS=\angle VUR =\angle UVR$

∴ In $\triangle UVR$

$\angle VUR= \angle RVU$

or,

$RV = UR = 2$ -------- $(ii)$

From $(i)$ and $(ii)$

$QS = VR = UR = 2$units

## (2) Comment(s)

KARTIK
()

correction : ∠QTS=∠VTR

By the way a good question.

Deepak
()