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Question
In the given figure, triangle PQR is right-angled at Q. S is the mid-point of side QR. Prove that QR2 = 4(PS2 – PQ2).
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Solution
Given: In triangle PQR, ∠PQR = 90° and S is the mid-point of QR.
To prove: QR2 = 4(PS2 – PQ2)
in right-angled ΔPQS, by Pythagoras theorem,
PQ2 + QS2 = PS2
⇒ QS2 = PS2 – PQ2 .......(i)
Since S is the mid-point of side QR,
∴ QS = `(QR)/2`
Substituting the value of QS in equation (i),
`((QR)/2)^2 = PS^2 - PQ^2`
`(QR^2)/4 = PS^2 - PQ^2`
QR2 = 4(PS2 – PQ)2
Hence proved.
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