ABCD is a quadrilateral such that `angleD=90^(@)`. A circle C of radius r touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm. Find r

Answer

To solve the problem, we need to find the radius \( r \) of the circle that touches all four sides of the quadrilateral \( ABCD \) with \( \angle D = 90^\circ \). Given the lengths \( BC = 38 \, \text{cm} \), \( CD = 25 \, \text{cm} \), and \( BP = 27 \, \text{cm} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Tangent Segments**:
Since the circle touches the sides of the quadrilateral at points \( P, Q, R, \) and \( S \), we know that the lengths of the tangent segments from a point to a circle are equal. Therefore, we can denote:
- \( BP = BQ = 27 \, \text{cm} \)
- Let \( CQ = CR \) (we will find this value).
- Let \( RD = RS \) (we will find this value).

2. **Calculate \( CQ \)**:
We know that:
\[ BC = BP + CQ \]
Substituting the known values:
\[ 38 = 27 + CQ \]
Solving for \( CQ \):
\[ CQ = 38 - 27 = 11 \, \text{cm} \]
Thus, \( CQ = CR = 11 \, \text{cm} \).

3. **Calculate \( RD \)**:
We know that:
\[ CD = CR + RD \]
Substituting the known values:
\[ 25 = 11 + RD \]
Solving for \( RD \):
\[ RD = 25 - 11 = 14 \, \text{cm} \]

4. **Identify the Relationship in Quadrilateral**:
In quadrilateral \( OSRD \) (where \( O \) is the center of the circle), we have:
- \( OS = OR = r \) (the radius of the circle).
- \( RD = 14 \, \text{cm} \).

5. **Recognize the Right Triangle**:
Since \( \angle D = 90^\circ \), triangle \( OSD \) is a right triangle with:
- \( OS \) as one leg,
- \( RD \) as the other leg,
- \( OD \) as the hypotenuse.

6. **Establish the Square**:
Since \( OS = OR \) and \( RD \) is perpendicular to \( OS \), quadrilateral \( OSRD \) forms a square. Therefore, all sides are equal:
\[ OS = RD = 14 \, \text{cm} \]

7. **Conclusion**:
Thus, the radius \( r \) of the circle is:
\[ r = 14 \, \text{cm} \]

### Final Answer:
The radius \( r \) of the circle is \( 14 \, \text{cm} \).