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In the figure given below, Delta PQR is ...

In the figure given below, `Delta PQR` is right-angled at Q and the points S and T trisect the side QR. Prove that `8PT^(2)=3PR^(2)+5PS^(2)`

Text Solution

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GIVEN ` A Delta PQR` in which `angle PQR=90^(@)`, S and T are the points of trisection of QR.
TO PROVE `8 TP^(2)=3PR^(2)+5PS^(2)`
PROOF Let `AQ=ST=TR=x`. Then,
`QS=x,QT=2x and QR=3x`.
From right triangle `PQS, PQT and PQR` by Pythagora's theorem we have,

`PS^(2)=PQ^(2)+QS^(2),PT^(2)=PQ^(2)+QT^(2) and PR^(2)= PQ^(2)+QR^(2)`
`:. 3PR^(2)+5PS^(2)-TP^(2)`
`= 3(PQ^(2)+QR^(2))+5(PQ^(2)+QS^(2))-8(PQ^(2)-QT^(2))`
`=3QR^(2)+5QS^(2)-8QT^(2)` ltbr `=3xx(3x)^(2)+5(x)^(2)-8xx(2x)^(2) [ :. QR=3x, QS= x and QT=2x]`
`=(27x+5x^(2)-x^(2))=0`
Thus, `3PR^(2)+5PS^(2)-8PT^(2)=0`
Hence, `8PT^(2)=3PR^(2)=3PR^(2)+5PS^(2)`
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