An isoceles right triangl ABC has an incircle with centre O, which touches the sides AB, BC, and AC . Find radius of circle

Options

A. `sqrt2-1` cm
B. `sqrt2-1/2` cm
C. `1-1/sqrt2` cm
D. none of these

Answer

To find the radius of the incircle of an isosceles right triangle ABC, we can follow these steps:

### Step 1: Understand the Triangle
Since triangle ABC is an isosceles right triangle, we can denote the lengths of the two equal sides (legs) as \( a \). The hypotenuse will then be \( a\sqrt{2} \).

### Step 2: Calculate the Area of the Triangle
The area \( A \) of a right triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
For triangle ABC, both the base and height are equal to \( a \):
\[ A = \frac{1}{2} \times a \times a = \frac{a^2}{2} \]

### Step 3: Calculate the Semi-Perimeter
The semi-perimeter \( s \) of triangle ABC is given by:
\[ s = \frac{\text{side}_1 + \text{side}_2 + \text{hypotenuse}}{2} = \frac{a + a + a\sqrt{2}}{2} = \frac{2a + a\sqrt{2}}{2} = a\left(1 + \frac{\sqrt{2}}{2}\right) \]

### Step 4: Use the Formula for the Radius of the Incircle
The radius \( r \) of the incircle can be calculated using the formula:
\[ r = \frac{A}{s} \]
Substituting the values we found:
\[ r = \frac{\frac{a^2}{2}}{a\left(1 + \frac{\sqrt{2}}{2}\right)} \]

### Step 5: Simplify the Expression
Now we simplify the expression for \( r \):
\[ r = \frac{a^2}{2a\left(1 + \frac{\sqrt{2}}{2}\right)} = \frac{a}{2\left(1 + \frac{\sqrt{2}}{2}\right)} \]
To simplify further, multiply the numerator and denominator by 2:
\[ r = \frac{2a}{4 + 2\sqrt{2}} = \frac{2a(2 - \sqrt{2})}{(4 + 2\sqrt{2})(2 - \sqrt{2})} \]
Using the difference of squares in the denominator:
\[ (4 + 2\sqrt{2})(2 - \sqrt{2}) = 8 - 4\sqrt{2} + 4 - 2 = 12 - 6\sqrt{2} \]
Thus, the radius simplifies to:
\[ r = \frac{2a(2 - \sqrt{2})}{12 - 6\sqrt{2}} \]

### Step 6: Final Result
Assuming \( a = 1 \) (for simplicity), we can substitute \( a \) into the formula:
\[ r = \frac{2(2 - \sqrt{2})}{12 - 6\sqrt{2}} \]

### Conclusion
The radius of the incircle of the isosceles right triangle ABC is given by the simplified expression above.

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