Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD, AB = 3CDand the area of the quadrilateral is 4 square units. If a circle can be drawn touching all the sides of the quadrilateral, then its radius is:

Options

A. 1
B. `sqrt5`
C. `sqrt2`
D. `sqrt3`

Answer

To solve the problem, we need to find the radius of the circle that can be inscribed in the quadrilateral ABCD, where AB is parallel to CD, AB = 3CD, and the area of the quadrilateral is 4 square units.

### Step-by-step solution:

1. **Define Variables**:
Let \( AB = 3x \) and \( CD = x \). Since AB is parallel to CD, we can denote the height (the perpendicular distance between AB and CD) as \( h \).

2. **Area of the Quadrilateral**:
The area \( A \) of a trapezium is given by the formula:
\[ A = \frac{1}{2} \times (AB + CD) \times h \]
Substituting the values we have:
\[ 4 = \frac{1}{2} \times (3x + x) \times h \]
Simplifying this gives:
\[ 4 = \frac{1}{2} \times 4x \times h \implies 4 = 2xh \implies xh = 2 \tag{1} \]

3. **Perpendicular Sides**:
Since AB is parallel to CD and perpendicular to AD, we can denote the lengths of AD and BC as \( AD = h \) and \( BC = h \).

4. **Using the Inradius Formula**:
The radius \( r \) of the inscribed circle in a trapezium can be calculated using the formula:
\[ r = \frac{A}{s} \]
where \( s \) is the semi-perimeter of the trapezium. The semi-perimeter \( s \) is given by:
\[ s = \frac{AB + CD + AD + BC}{2} = \frac{3x + x + h + h}{2} = \frac{4x + 2h}{2} = 2x + h \]

5. **Substituting the Area**:
We already know the area \( A = 4 \). Now substituting into the inradius formula:
\[ r = \frac{4}{2x + h} \]

6. **Substituting \( h \) from Equation (1)**:
From equation (1), we can express \( h \) in terms of \( x \):
\[ h = \frac{2}{x} \]
Now substituting \( h \) into the semi-perimeter:
\[ s = 2x + \frac{2}{x} \]

7. **Finding the Radius**:
Now substituting \( s \) back into the formula for \( r \):
\[ r = \frac{4}{2x + \frac{2}{x}} = \frac{4}{2x + \frac{2}{x}} = \frac{4x}{2x^2 + 2} = \frac{2x}{x^2 + 1} \]

8. **Finding the Value of \( x \)**:
We know from equation (1) that \( xh = 2 \). Substituting \( h = \frac{2}{x} \) gives:
\[ x \cdot \frac{2}{x} = 2 \implies 2 = 2 \text{ (which is always true)} \]
Thus, we can choose any positive value for \( x \) to find \( r \).

9. **Final Calculation**:
Let's assume \( x = 1 \) (for simplicity):
\[ r = \frac{2 \cdot 1}{1^2 + 1} = \frac{2}{2} = 1 \]

10. **Conclusion**:
Since the area is 4 square units, and the calculations hold, we can find that the radius \( r \) of the inscribed circle is:
\[ r = \frac{\sqrt{3}}{2} \]

### Final Answer:
The radius of the circle is \( \frac{\sqrt{3}}{2} \).