Question

f $a + b + c = 2s$ , then $\dfrac{{{{\left( {s - a} \right)}^2} + {{\left( {s - b} \right)}^2} + {{\left( {s - c} \right)}^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}}$ is equal to
A. ${a^2} + {b^2} + {c^2}$
B. 0
C. 1
D. 2

Answer 1

Hint: Expand the equation $\dfrac{{{{\left( {s - a} \right)}^2} + {{\left( {s - b} \right)}^2} + {{\left( {s - c} \right)}^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}}$. And substitute the value of ‘s’ as $a + b + c = 2s$ where ever necessary in the solution to get the value.


Complete step-by-step solution:
We are given $a + b + c = 2s$
We have to find the value of $\dfrac{{{{\left( {s - a} \right)}^2} + {{\left( {s - b} \right)}^2} + {{\left( {s - c} \right)}^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}}$
Expand the above polynomial
$ = \dfrac{{{{\left( {s - a} \right)}^2} + {{\left( {s - b} \right)}^2} + {{\left( {s - c} \right)}^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}}$
Use the formula of ${\left( {a - b} \right)^2}$
$
   = \dfrac{{\left( {{s^2} - 2sa + {a^2}} \right) + \left( {{s^2} - 2sb + {b^2}} \right) + \left( {{s^2} - 2sc + {c^2}} \right) + {s^2}}}{{{a^2} + {b^2} + {c^2}}} \\
  \left( {\because {{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right) \\
   = \dfrac{{{s^2} - 2sa + {a^2} + {s^2} - 2sb + {b^2} + {s^2} - 2sc + {c^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}} \\
 $
And then put all the similar terms together
$ = \dfrac{{4{s^2} - 2sa - 2sb - 2sc + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}}$
Take out ‘2s’ common
$ = \dfrac{{4{s^2} - 2s\left( {a + b + c} \right) + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}}$
As given in the question, $a + b + c = 2s$, substitute the value 2s in the above equation
After substituting the value, the above equation becomes
$
   \to \dfrac{{4{s^2} - 2s\left( {a + b + c} \right) + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}} \\
   = \dfrac{{4{s^2} - 2s\left( {2s} \right) + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}} \\
  \left( {\because a + b + c = 2s} \right) \\
   = \dfrac{{4{s^2} - 4{s^2} + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}} \\
   = \dfrac{{{a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}} \\
   = 1 \\
 $
The value of $\dfrac{{{{\left( {s - a} \right)}^2} + {{\left( {s - b} \right)}^2} + {{\left( {s - c} \right)}^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}}$ is 1.
Therefore, from among the options given in the question option c is correct, which is 1.
Additional information: Substitution method is used here to solve the problem. The method of solving “by substitution” is solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, “substituting” for the chosen variable and solving for the other variable.


Note: Another approach to the above problem
We have to find the value of $\dfrac{{{{\left( {s - a} \right)}^2} + {{\left( {s - b} \right)}^2} + {{\left( {s - c} \right)}^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}}$
Expand the above polynomial
$ = \dfrac{{{{\left( {s - a} \right)}^2} + {{\left( {s - b} \right)}^2} + {{\left( {s - c} \right)}^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}}$
Use the formula of ${\left( {a - b} \right)^2}$
$
   = \dfrac{{\left( {{s^2} - 2sa + {a^2}} \right) + \left( {{s^2} - 2sb + {b^2}} \right) + \left( {{s^2} - 2sc + {c^2}} \right) + {s^2}}}{{{a^2} + {b^2} + {c^2}}} \\
  \left( {\because {{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right) \\
   = \dfrac{{{s^2} - 2sa + {a^2} + {s^2} - 2sb + {b^2} + {s^2} - 2sc + {c^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}} \\
 $
And then put all the similar terms together
$ = \dfrac{{4{s^2} - 2sa - 2sb - 2sc + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}}$
Take out ‘2s’ common
$ = \dfrac{{4{s^2} - 2s\left( {a + b + c} \right) + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}}$
As given in the question, $a + b + c = 2s$, substitute the value of ‘2s’ in the above equation
After substituting the value, the above equation becomes
$
   \to \dfrac{{{{\left( {2s} \right)}^2} - 2s\left( {a + b + c} \right) + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}} \\
   = \dfrac{{{{\left( {a + b + c} \right)}^2} - \left( {a + b + c} \right)\left( {a + b + c} \right) + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}} \\
  \left( {\because a + b + c = 2s} \right) \\
   = \dfrac{{{{\left( {a + b + c} \right)}^2} - {{\left( {a + b + c} \right)}^2} + {a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}} \\
   = \dfrac{{{a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2}}} \\
   = 1 \\
 $
The value of $\dfrac{{{{\left( {s - a} \right)}^2} + {{\left( {s - b} \right)}^2} + {{\left( {s - c} \right)}^2} + {s^2}}}{{{a^2} + {b^2} + {c^2}}}$ is 1.