PQR is a right angled triangle at Q with QR =12 a=cm and PQ=5cm. A circle with centre O is inscribed in `trianglePQR.` Find the radius of the circle.

Answer

To find the radius of the inscribed circle in triangle PQR, we will follow these steps:

### Step 1: Identify the sides of the triangle
Given:
- QR = 12 cm (one leg of the triangle)
- PQ = 5 cm (the other leg of the triangle)

Since triangle PQR is a right-angled triangle at Q, we can denote:
- PQ = a = 5 cm
- QR = b = 12 cm

### Step 2: Use the Pythagorean theorem to find PR
According to the Pythagorean theorem:
\[ PR^2 = PQ^2 + QR^2 \]
Substituting the values:
\[ PR^2 = 5^2 + 12^2 \]
\[ PR^2 = 25 + 144 \]
\[ PR^2 = 169 \]
Taking the square root:
\[ PR = \sqrt{169} = 13 \text{ cm} \]

### Step 3: Calculate the semi-perimeter (s) of triangle PQR
The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} \]
Where \( c = PR \):
\[ s = \frac{5 + 12 + 13}{2} \]
\[ s = \frac{30}{2} = 15 \text{ cm} \]

### Step 4: Use the formula for the radius (r) of the inscribed circle
The radius \( r \) of the inscribed circle in a triangle can be calculated using the formula:
\[ r = \frac{A}{s} \]
Where \( A \) is the area of the triangle.

### Step 5: Calculate the area (A) of triangle PQR
The area \( A \) of a right-angled triangle can be calculated as:
\[ A = \frac{1}{2} \times base \times height \]
Using PQ as the base and QR as the height:
\[ A = \frac{1}{2} \times 5 \times 12 \]
\[ A = \frac{1}{2} \times 60 = 30 \text{ cm}^2 \]

### Step 6: Substitute the values into the radius formula
Now substituting the area and semi-perimeter into the radius formula:
\[ r = \frac{A}{s} = \frac{30}{15} = 2 \text{ cm} \]

### Final Answer
The radius of the inscribed circle is \( 2 \text{ cm} \).

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