Triangle PQR circumscribes a circle with centre O and radius r cm such that `/_PQR = 90^@`. If `PQ = 3 cm, QR = 4 cm,` then the value of r is :

Options

A. 2
B. 1.5
C. 2.5
D. 1

Answer

To find the radius \( r \) of the incircle of triangle \( PQR \) where \( \angle PQR = 90^\circ \), \( PQ = 3 \, \text{cm} \), and \( QR = 4 \, \text{cm} \), we can follow these steps:

### Step 1: Identify the sides of the triangle
Since \( \angle PQR \) is a right angle, we can identify the sides:
- \( PQ = 3 \, \text{cm} \) (one leg)
- \( QR = 4 \, \text{cm} \) (other leg)

To find the hypotenuse \( PR \), we can use the Pythagorean theorem:
\[ PR = \sqrt{PQ^2 + QR^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{cm} \]

### Step 2: Calculate the area of triangle \( PQR \)
The area \( \Delta \) of a right triangle can be calculated using the formula:
\[ \Delta = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take \( PQ \) as the base and \( QR \) as the height:
\[ \Delta = \frac{1}{2} \times 3 \times 4 = \frac{12}{2} = 6 \, \text{cm}^2 \]

### Step 3: Calculate the semi-perimeter \( s \)
The semi-perimeter \( s \) of triangle \( PQR \) is given by:
\[ s = \frac{a + b + c}{2} \]
where \( a = PQ \), \( b = QR \), and \( c = PR \):
\[ s = \frac{3 + 4 + 5}{2} = \frac{12}{2} = 6 \, \text{cm} \]

### Step 4: Calculate the radius \( r \) of the incircle
The radius \( r \) of the incircle can be calculated using the formula:
\[ r = \frac{\Delta}{s} \]
Substituting the values we found:
\[ r = \frac{6}{6} = 1 \, \text{cm} \]

### Final Answer
Thus, the value of \( r \) is \( 1 \, \text{cm} \).

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