If a + b + c = 2s, then the value of (s - a)² + (s - b)² + (s - c)² + s² will be-

1s² - a² - b² - c²
2s² + a² + b² + c²
3a² + b² + c²
44s² - a² - b² - c²

Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(s - a)}^{2} + {(s - b)}^{2} + {(s - c)}^{2} + s^{2} \\ = (s^{2} + a^{2} - 2 s a) + (s^{2} + b^{2} - 2 s b) + \\ (s^{2} + c^{2} - 2 s c) + s^{2} \\ = 4 s^{2} + (a^{2} + b^{2} + c^{2}) - 2 s (a + b + c) \\ = (2 s)^{2} + (a^{2} + b^{2} + c^{2}) - \\ (a + b + c) (a + b + c) \\ = (a + b + c)^{2} + (a^{2} + b^{2} + c^{2}) - \\ (a + b + c)^{2} \\ = a^{2} + b^{2} + c^{2} \end{aligned}

$\eqalign{ & {\left( {s - a} \right)^2} + {\left( {s - b} \right)^2} + {\left( {s - c} \right)^2} + {s^2} \cr & = \left( {{s^2} + {a^2} - 2sa} \right) + \left( {{s^2} + {b^2} - 2sb} \right) + \cr & \,\,\,\,\,\,\left( {{s^2} + {c^2} - 2sc} \right) + {s^2} \cr & = 4{s^2} + ({a^2} + {b^2} + {c^2}) - 2s(a + b + c) \cr & = {(2s)^2} + ({a^2} + {b^2} + {c^2}) - \cr & \,\,\,\,\,\,(a + b + c)(a + b + c) \cr & = {(a + b + c)^2} + \left( {{a^2} + {b^2} + {c^2}} \right) - \cr & \,\,\,\,\,\,{(a + b + c)^2} \cr & = {a^2} + {b^2} + {c^2} \cr}$