Question

IN ADJOINING FIGURE ,SEG PS IS THE MEDIAN OF TRIANGLE PQR AND PTPERPUDICULAR TO QR . PROVE THAT ,
1}. PR2 =PS2+QR*ST+[QR/2]2

Answer 1

Step-by-step explanation:

Seg PS is the median of ∆PQR.......(given)

QS=SR=1/2QR ...(S is the midpoint of side QR)..(1)

In∆PTS,

PTS=90°

By Pythagoras theorem,

PS2=PT2+TS2......(2)

In ∆ PTR,PTR=90° ....(given)

By Pythagoras theorem,

PR2=PT2+TR2

PR2=PT2+(TS+SR)2....(T-S-R)

PR2=PT2+TS2+2ST*SR+SR2

....[(a+b)2=a2+2ab+b2)]

PR2=(PT2+TS2)+2ST*SR+SR2

PR2=PS2+2ST*(QR/2)+(QR/2)2

...( From 1 and 2)

PR2=PS2+QR*ST+[QR/2]2

....( Hence proved)