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In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
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In ∆PQR, point S is the midpoint of side QR. [Given]
∴ seg PS is the median.
∴ PQ2 + PR2 = 2 PS2 + 2 SR2 [Apollonius theorem] 

∴ 112 + 172 = 2 (13)2 + 2 SR2
∴ 121 + 289 = 2 (169)+ 2 SR2
∴ 410 = 338+ 2 SR2
∴ 2 SR2 = 410 – 338
∴ 2 SR2 = 72
∴ SR2 = \(72\over2\) = 36
∴ SR = \(\sqrt{36}\) [Taking square root of both sides]
= 6 units Now, QR = 2 SR [S is the midpoint of QR]
= 2 × 6


∴ QR = 12 units

{no need to write just for explaination }
Statement- “the sum of squares of any of the two sides of a triangle equals to twice its square on half of the third side, along with the twice of its square on the median bisecting the third side” 
For deep explaination . Visit  This link https://edu.brainiak.in/post/apollonius-theorem-with-proof
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