In triangle ABC, of `r_(1)= 2r_(2)=3r_(3)` Then a:b is equal :-

Options

A. `4/5`
B. `5/4`
C. `7/4`
D. `4/7`

Answer

To solve the problem where \( r_1 = 2r_2 = 3r_3 \) in triangle ABC and find the ratio \( a:b \), we can follow these steps:

### Step 1: Understand the relationship between the inradii and the sides of the triangle

We know that the inradii \( r_1, r_2, r_3 \) are given by the formulas:
\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \]
where \( \Delta \) is the area of the triangle and \( s \) is the semi-perimeter defined as \( s = \frac{a + b + c}{2} \).

### Step 2: Set up the equations based on the given ratios

Given \( r_1 = 2r_2 \) and \( r_1 = 3r_3 \), we can write:
\[ \frac{\Delta}{s - a} = 2 \cdot \frac{\Delta}{s - b} \quad \text{(1)} \]
\[ \frac{\Delta}{s - a} = 3 \cdot \frac{\Delta}{s - c} \quad \text{(2)} \]

### Step 3: Simplify the equations

From equation (1):
\[ \frac{1}{s - a} = \frac{2}{s - b} \]
Cross-multiplying gives:
\[ s - b = 2(s - a) \implies s - b = 2s - 2a \implies s - 2s + 2a - b = 0 \implies 2a - b - s = 0 \implies s = 2a - b \quad \text{(3)} \]

From equation (2):
\[ \frac{1}{s - a} = \frac{3}{s - c} \]
Cross-multiplying gives:
\[ s - c = 3(s - a) \implies s - c = 3s - 3a \implies -c + 3a - 2s = 0 \implies c = 3a - 2s \quad \text{(4)} \]

### Step 4: Substitute \( s \) from equation (3) into equation (4)

Substituting \( s = 2a - b \) into equation (4):
\[ c = 3a - 2(2a - b) = 3a - 4a + 2b = -a + 2b \quad \text{(5)} \]

### Step 5: Use the semi-perimeter to relate \( a, b, c \)

The semi-perimeter \( s \) can also be expressed as:
\[ s = \frac{a + b + c}{2} \]
Substituting \( c \) from equation (5):
\[ s = \frac{a + b + (-a + 2b)}{2} = \frac{3b}{2} \]
Setting this equal to equation (3):
\[ 2a - b = \frac{3b}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 4a - 2b = 3b \implies 4a = 5b \implies \frac{a}{b} = \frac{5}{4} \]

### Final Result

Thus, the ratio \( a:b \) is:
\[ \boxed{5:4} \]