In a `Delta ABC` AB = 13 , BC = 14 , CA = 15 find the radius of excentre .

Answer

To find the radius of the excentre in triangle ABC where AB = 13, BC = 14, and CA = 15, we will follow these steps:

### Step 1: Calculate the semi-perimeter (s)
The semi-perimeter \( s \) of a triangle is given by the formula:
\[ s = \frac{a + b + c}{2} \]
where \( a \), \( b \), and \( c \) are the lengths of the sides of the triangle.

For triangle ABC:
- \( a = BC = 14 \)
- \( b = CA = 15 \)
- \( c = AB = 13 \)

Calculating the semi-perimeter:
\[ s = \frac{14 + 15 + 13}{2} = \frac{42}{2} = 21 \]

### Step 2: Calculate the area (Δ) of the triangle using Heron's formula
Heron's formula for the area \( Δ \) of a triangle is given by:
\[ Δ = \sqrt{s(s-a)(s-b)(s-c)} \]

Substituting the values:
- \( s = 21 \)
- \( s - a = 21 - 14 = 7 \)
- \( s - b = 21 - 15 = 6 \)
- \( s - c = 21 - 13 = 8 \)

Now, substituting these into Heron's formula:
\[ Δ = \sqrt{21 \times 7 \times 6 \times 8} \]

Calculating:
\[ Δ = \sqrt{21 \times 7 \times 6 \times 8} = \sqrt{21 \times 336} = \sqrt{7056} \]
Calculating \( \sqrt{7056} \):
\[ Δ = 84 \]

### Step 3: Calculate the radius of the excentres (R_a, R_b, R_c)
The radius of the excentres can be calculated using the formulas:
\[ R_a = \frac{Δ}{s-a}, \quad R_b = \frac{Δ}{s-b}, \quad R_c = \frac{Δ}{s-c} \]

Calculating \( R_a \):
\[ R_a = \frac{84}{s - a} = \frac{84}{21 - 14} = \frac{84}{7} = 12 \]

Calculating \( R_b \):
\[ R_b = \frac{84}{s - b} = \frac{84}{21 - 15} = \frac{84}{6} = 14 \]

Calculating \( R_c \):
\[ R_c = \frac{84}{s - c} = \frac{84}{21 - 13} = \frac{84}{8} = 10.5 \]

### Conclusion
The radius of the excentres are:
- \( R_a = 12 \)
- \( R_b = 14 \)
- \( R_c = 10.5 \)