In the given figure, $PQR$ is a right triangle right-angled at $Q$. $S$ and $T$ trisect $QR$. Prove that $8 PT^2 = 3 PR^2 + 5 PS^2$.
Step-by-Step Solution
Key Concept: Trisection: Let $QS = x, QT = 2x, QR = 3x$. In right $\Delta s$: $PS^2 = PQ^2 + x^2$, $PT^2 = PQ^2 + 4x^2$, $PR^2 = PQ^2 + 9x^2$. Calculate $3PR^2 + 5PS^2 = 3(PQ^2+9x^2) + 5(PQ^2+x^2) = 8PQ^2 + 32x^2 = 8(PQ^2 + 4x^2) = 8PT^2$.
Let $QS = x, QT = 2x, QR = 3x$. By Pythagoras theorem:
$PS^2 = PQ^2 + x^2$, $PT^2 = PQ^2 + 4x^2$, $PR^2 = PQ^2 + 9x^2$. [1.0 Mark]
RHS $= 3(PQ^2 + 9x^2) + 5(PQ^2 + x^2) = 3PQ^2 + 27x^2 + 5PQ^2 + 5x^2 = 8PQ^2 + 32x^2$. [1.0 Mark]
$= 8(PQ^2 + 4x^2) = 8 PT^2 = $ LHS. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Expressing $PS^2, PT^2, PR^2$ in terms of $PQ$ and $x$: 1.0 Mark
Expanding RHS $= 8PQ^2 + 32x^2$: 1.0 Mark
Factoring into $8(PQ^2 + 4x^2) = 8PT^2$: 1.0 Mark
Correct Answer: