Question

2s=a+b+c
Prove that:
${(s - a)}^{2} + {(s - b)}^{2} + {(s - c)}^{2} + {s}^{2} \\ = {a}^{2} + {b}^{2} + {c}^{2}$
...Plz show the required steps...​

Answer 1

Step-by-step explanation:

Given Equation is: 2s = a + b + c  ------ (i)

LHS:

(s - a)² + (s - b)² + (s - c)² + s²

⇒ (s² + a² - 2as) + (s² + b² - 2bs) + (s² + c² - 2cs) + s²

⇒ s² + a² - 2as + s² + b² - 2bs + s² + c² - 2cs + s²

⇒ 4s² + (a² + b² + c²) - 2as - 2bs - 2cs

⇒ 4s² + (a² + b² + c²) - 2s(a + b + c)

⇒ (2s)² + (a² + b² + c²) - (a + b + c)(a + b + c)

⇒ (a + b + c)² + (a² + b² + c²) - (a + b + c)²

⇒ a² + b² + c²

RHS

Hope it helps!