In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

In ∆PQR, point S is the midpoint of side QR. [Given]

∴ seg PS is the median.

∴ PQ^{2} + PR^{2} = 2 PS^{2} + 2 SR^{2} [Apollonius theorem]

∴ 11^{2} + 17^{2} = 2 (13)^{2} + 2 SR^{2}

∴ 121 + 289 = 2 (169)+ 2 SR^{2}

∴ 410 = 338+ 2 SR^{2}

∴ 2 SR^{2} = 410 – 338

∴ 2 SR^{2} = 72

∴ SR^{2} = \(72\over2\) = 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`