[2 units] Through T, the mid-point of the side QR of a \triangle PQR , a straight line is drawn to meet PQ

For Competitive Exams

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

$2\sqrt{2}units$

$\sqrt{2} units$

2 units

5 units

Explanation:

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

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

$\therefore$QS=VR -------- (i)

Now

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

$\therefore$ In $\triangle$UVR

$\angle$VUR=$\angle$RVU

or,

RV=UR=2 -------- (ii)

From (i) and (ii)

QS=VR=UR=2units

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