Geometry & Mensuration
Aptitude

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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

Image for Geometry and Mensuration, Aptitude:2190-1

 A.

\(2\sqrt{2}\) units

 B.

\(\sqrt{2}\) units

 C.

\(2\)units

 D.

cannot be determined

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Solution:
Option(C) is correct

We have \(QT = TR\) and \(PU = PS. UR = 2\) units

Answer image for Geometry and Mensuration, Aptitude:2190-1

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
 ()

Thank you Kartik, correction made.