The distance of incentre of the right-angled triangle ABC (right angled at A) from B and C are `sqrt10 and sqrt5`, respectively. The perimeter of the triangle is _____

Explanation

To find the perimeter of the right-angled triangle ABC, where the distances from the incenter to points B and C are given as \( \sqrt{10} \) and \( \sqrt{5} \) respectively, we can follow these steps:

### Step 1: Define the triangle and the incenter
Let \( A \) be the right angle of triangle \( ABC \), with \( B \) and \( C \) being the other two vertices. The incenter \( I \) is the point where the angle bisectors of the triangle meet, and it is equidistant from all sides of the triangle.

### Step 2: Set up the distances from the incenter
We know:
- The distance from \( I \) to \( B \) (denote it as \( r_B \)) is \( \sqrt{10} \).
- The distance from \( I \) to \( C \) (denote it as \( r_C \)) is \( \sqrt{5} \).

### Step 3: Relate the distances to the sides of the triangle
In a right-angled triangle, the distances from the incenter to the vertices can be expressed in terms of the inradius \( r \) and the lengths of the sides opposite to those vertices. For triangle \( ABC \):
- Let \( a \) be the length of side \( BC \),
- Let \( b \) be the length of side \( AC \),
- Let \( c \) be the length of side \( AB \).

Using the properties of the incenter, we can express:
- \( r_B = \frac{r}{\sin(\frac{B}{2})} \)
- \( r_C = \frac{r}{\sin(\frac{C}{2})} \)

### Step 4: Use the sine rule
Since \( B + C = 90^\circ \), we have:
- \( \sin(\frac{B}{2}) = \cos(\frac{C}{2}) \)
- \( \sin(\frac{C}{2}) = \cos(\frac{B}{2}) \)

This gives us the relationships:
\[ \sqrt{10} = \frac{r}{\cos(\frac{C}{2})}, \quad \sqrt{5} = \frac{r}{\cos(\frac{B}{2})} \]

### Step 5: Find the ratios
From the above equations, we can set up the ratio:
\[ \frac{\sqrt{10}}{\sqrt{5}} = \frac{\cos(\frac{C}{2})}{\cos(\frac{B}{2})} \]
This simplifies to:
\[ \frac{\sqrt{2}}{1} = \frac{\cos(\frac{C}{2})}{\cos(\frac{B}{2})} \]

### Step 6: Solve for the angles
Let \( \cos(\frac{B}{2}) = k \), then \( \cos(\frac{C}{2}) = \sqrt{2} k \). Using the identity \( \cos^2(\frac{B}{2}) + \cos^2(\frac{C}{2}) = 1 \):
\[ k^2 + 2k^2 = 1 \implies 3k^2 = 1 \implies k^2 = \frac{1}{3} \implies k = \frac{1}{\sqrt{3}} \]

Thus:
\[ \cos(\frac{B}{2}) = \frac{1}{\sqrt{3}}, \quad \cos(\frac{C}{2}) = \frac{\sqrt{2}}{\sqrt{3}} \]

### Step 7: Find the inradius
Using the distances:
\[ r = \sqrt{10} \cos(\frac{C}{2} ) = \sqrt{10} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{20}}{\sqrt{3}} = \frac{2\sqrt{5}}{\sqrt{3}} \]

### Step 8: Calculate the sides
Using the relationships:
\[ x = 2r \quad \text{and} \quad y = 3r \]
Substituting \( r \):
\[ x = 2 \cdot \frac{2\sqrt{5}}{\sqrt{3}} = \frac{4\sqrt{5}}{\sqrt{3}}, \quad y = 3 \cdot \frac{2\sqrt{5}}{\sqrt{3}} = \frac{6\sqrt{5}}{\sqrt{3}} \]

### Step 9: Find the perimeter
The perimeter \( P \) of triangle \( ABC \) is given by:
\[ P = a + b + c = x + y + r = \frac{4\sqrt{5}}{\sqrt{3}} + \frac{6\sqrt{5}}{\sqrt{3}} + \frac{2\sqrt{5}}{\sqrt{3}} = \frac{12\sqrt{5}}{\sqrt{3}} \]

### Step 10: Simplify the perimeter
Thus, the perimeter can be simplified to:
\[ P = 12 \text{ units} \]

### Final Answer:
The perimeter of triangle ABC is \( 12 \) units.

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