Value of `1/(r_(1)^2)'+ 1/(r_(2)^2)+ 1/(r_(3)^2)+ 1/(r_()^2)` is :

Options

A. 0
B. `(a^2+b^2+c^2)/(Delta^2)`
C. `(Delta^2)/(a^2+b^2+c^2)`
D. `(a^2+b^2+c^2)/(Delta)`

Answer

To solve the problem, we need to find the value of

\[ \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} \]

where \( r_1, r_2, r_3, \) and \( r \) are defined in terms of the area \( \Delta \) of a triangle and its semi-perimeter \( s \).

### Step 1: Define the values of \( r_1, r_2, r_3, \) and \( r \)

We know that:
- \( r_1 = \frac{\Delta}{s - a} \)
- \( r_2 = \frac{\Delta}{s - b} \)
- \( r_3 = \frac{\Delta}{s - c} \)
- \( r = \frac{\Delta}{s} \)

### Step 2: Substitute these values into the equation

Now, substituting these values into the equation, we have:

\[ \frac{1}{r_1^2} = \frac{(s - a)^2}{\Delta^2}, \quad \frac{1}{r_2^2} = \frac{(s - b)^2}{\Delta^2}, \quad \frac{1}{r_3^2} = \frac{(s - c)^2}{\Delta^2}, \quad \frac{1}{r^2} = \frac{s^2}{\Delta^2} \]

Thus, we can rewrite the equation as:

\[ \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{(s - a)^2 + (s - b)^2 + (s - c)^2 + s^2}{\Delta^2} \]

### Step 3: Simplify the numerator

Now, we need to simplify the numerator:

\[ (s - a)^2 + (s - b)^2 + (s - c)^2 + s^2 \]

Expanding each term, we get:

\[ (s^2 - 2as + a^2) + (s^2 - 2bs + b^2) + (s^2 - 2cs + c^2) + s^2 \]

Combining like terms:

\[ 4s^2 - 2s(a + b + c) + (a^2 + b^2 + c^2) \]

### Step 4: Substitute \( a + b + c \)

Since \( a + b + c = 2s \), we can substitute this into the equation:

\[ 4s^2 - 2s(2s) + (a^2 + b^2 + c^2) = 4s^2 - 4s^2 + (a^2 + b^2 + c^2) = a^2 + b^2 + c^2 \]

### Step 5: Final result

Thus, we can conclude that:

\[ \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{a^2 + b^2 + c^2}{\Delta^2} \]

### Final Answer

The value is:

\[ \frac{a^2 + b^2 + c^2}{\Delta^2} \]