The radii `r_1, r_2,r_3` of the escribed circles of the triangle `A B C` are in H.P. If the area of the triangle is `24 c m^2a n d` its perimeter is 24cm, then the length of its largest side is 10 (b) 9 (c) 8 (d) none of these

Options

A. The length of longest side of triangle ABC is equal to 10
B. The radius of circle inscribed in triangle ABC is equal to 4.
C. The circumradius of triangle ABC is equal to 5.
D. The sides of triangle ABC are in A.P.

Explanation

To solve the problem step by step, we will use the properties of the triangle and the given conditions.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We are given that the radii \( r_1, r_2, r_3 \) of the escribed circles of triangle \( ABC \) are in Harmonic Progression (H.P.). We also know the area of the triangle is \( 24 \, \text{cm}^2 \) and its perimeter is \( 24 \, \text{cm} \). We need to find the length of the largest side.

2. **Using the Relationship of Escribed Circles:**
The radius of the escribed circles can be expressed in terms of the area \( \Delta \) and the semi-perimeter \( s \):
\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \]
where \( a, b, c \) are the sides of the triangle opposite to vertices \( A, B, C \) respectively, and \( s = \frac{a + b + c}{2} \).

3. **Condition of H.P.:**
Since \( r_1, r_2, r_3 \) are in H.P., the reciprocals \( \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3} \) are in Arithmetic Progression (A.P.). This gives us:
\[ \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3} \text{ are in A.P.} \]
Therefore:
\[ 2 \cdot \frac{1}{r_2} = \frac{1}{r_1} + \frac{1}{r_3} \]
Substituting the expressions for \( r_1, r_2, r_3 \):
\[ 2 \cdot \frac{s - b}{\Delta} = \frac{s - a}{\Delta} + \frac{s - c}{\Delta} \]
Simplifying gives:
\[ 2(s - b) = (s - a) + (s - c) \]
Rearranging leads to:
\[ 2s - 2b = 2s - a - c \implies a + c = 2b \]

4. **Using the Perimeter:**
Given the perimeter \( a + b + c = 24 \):
\[ a + b + c = 24 \implies (2b) + b = 24 \implies 3b = 24 \implies b = 8 \]
Now substituting \( b = 8 \) into \( a + c = 2b \):
\[ a + c = 2 \cdot 8 = 16 \]

5. **Finding the Sides:**
We now have:
\[ a + c = 16 \quad \text{and} \quad b = 8 \]
Thus, we can express \( c \) in terms of \( a \):
\[ c = 16 - a \]

6. **Using the Area Formula:**
The area of the triangle is given by:
\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]
where \( s = \frac{24}{2} = 12 \):
\[ \Delta = 24 = \sqrt{12(12-a)(12-8)(12-(16-a))} \]
Simplifying gives:
\[ 24 = \sqrt{12(12-a)(4)(a-4)} \]
Squaring both sides:
\[ 576 = 48(12-a)(a-4) \]
Dividing by 48:
\[ 12 = (12-a)(a-4) \]
Expanding and rearranging:
\[ 12 = 12a - a^2 - 48 + 4a \implies a^2 - 16a + 60 = 0 \]

7. **Solving the Quadratic Equation:**
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 60}}{2 \cdot 1} \]
\[ = \frac{16 \pm \sqrt{256 - 240}}{2} = \frac{16 \pm \sqrt{16}}{2} = \frac{16 \pm 4}{2} \]
This gives us:
\[ a = 10 \quad \text{or} \quad a = 6 \]

8. **Finding the Sides:**
If \( a = 10 \), then \( c = 16 - 10 = 6 \) and \( b = 8 \). The sides are \( 10, 8, 6 \).

9. **Conclusion:**
The largest side of the triangle is \( 10 \, \text{cm} \).

### Final Answer:
The length of the largest side is **10 cm**.