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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). - Geometry Mathematics 2

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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).

Theorem

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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Converse of Pythagoras Theorem
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