The inradius in a right angled triangle with integer sides is r
If r = 5, the greatest area (in sq. units) is

Options

A. 150
B. 210
C. 330
D. 450

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the relationship between inradius and sides of the triangle
In a right-angled triangle, the inradius \( r \) can be expressed in terms of the sides \( a \), \( b \), and \( c \) (where \( c \) is the hypotenuse) as:
\[ r = \frac{a + b - c}{2} \]
Given \( r = 5 \), we can substitute this into the equation:
\[ 10 = a + b - c \quad \text{(Equation 1)} \]

### Step 2: Use the Pythagorean theorem
For a right-angled triangle, the relationship between the sides is given by the Pythagorean theorem:
\[ c^2 = a^2 + b^2 \]
From this, we can express \( c \) as:
\[ c = \sqrt{a^2 + b^2} \quad \text{(Equation 2)} \]

### Step 3: Substitute \( c \) into Equation 1
Substituting Equation 2 into Equation 1 gives:
\[ 10 = a + b - \sqrt{a^2 + b^2} \]
Rearranging this, we have:
\[ \sqrt{a^2 + b^2} = a + b - 10 \]

### Step 4: Square both sides
Squaring both sides to eliminate the square root:
\[ a^2 + b^2 = (a + b - 10)^2 \]
Expanding the right side:
\[ a^2 + b^2 = a^2 + b^2 + 100 - 20a - 20b + 2ab \]
Cancelling \( a^2 + b^2 \) from both sides:
\[ 0 = 100 - 20a - 20b + 2ab \]
Rearranging gives:
\[ 2ab - 20a - 20b + 100 = 0 \]
Dividing the entire equation by 2:
\[ ab - 10a - 10b + 50 = 0 \quad \text{(Equation 3)} \]

### Step 5: Find integer solutions for \( a \) and \( b \)
We can rearrange Equation 3 to find:
\[ ab - 10a - 10b = -50 \]
This can be factored as:
\[ (a - 10)(b - 10) = 50 \]
Now we need to find pairs of integers \( (a - 10) \) and \( (b - 10) \) that multiply to 50.

The pairs of factors of 50 are:
- (1, 50)
- (2, 25)
- (5, 10)
- (10, 5)
- (25, 2)
- (50, 1)

### Step 6: Calculate corresponding \( a \) and \( b \)
Using the factor pairs:
1. \( (1, 50) \) gives \( (11, 60) \)
2. \( (2, 25) \) gives \( (12, 35) \)
3. \( (5, 10) \) gives \( (15, 20) \)

### Step 7: Calculate the area for each pair
The area \( A \) of a right-angled triangle is given by:
\[ A = \frac{1}{2} \times a \times b \]

Calculating the area for each pair:
1. For \( (11, 60) \):
\[ A = \frac{1}{2} \times 11 \times 60 = 330 \text{ sq. units} \]
2. For \( (12, 35) \):
\[ A = \frac{1}{2} \times 12 \times 35 = 210 \text{ sq. units} \]
3. For \( (15, 20) \):
\[ A = \frac{1}{2} \times 15 \times 20 = 150 \text{ sq. units} \]

### Step 8: Determine the greatest area
The greatest area among the calculated areas is:
\[ \text{Greatest Area} = 330 \text{ sq. units} \]

### Final Answer
The greatest area of the right-angled triangle with integer sides and an inradius of 5 is:
\[ \boxed{330 \text{ sq. units}} \]