Let CD=x
then AB=2CD=2x.
Let r be the radius of the circle inscribed in the quadrilateral ABCD.
Given: Area of quadrilateral ABCD=18 and ABis|| to CD.
⇒(1/2)(x+2x).2r=18
⇒3xr=18
⇒xr=6-----(1)
OP=OM=PD=OQ=AM=r
⇒PC=x−r and MB=2x−r
Let ∠PCO = angleOCQ = θ then from right-angled ΔOPC
tanθ=OP/CP=r/(x−r)(2)
CD∥AB
∴∠PCB=∠QOM=2θ
Step 2:
∠CBA=180∘−2θ
∠OBM=90∘−θ
⇒ From ΔOMB. ,tan(90∘−θ)=OM/MB=r/(2x−r)
From right angled ΔOBM
tanθ=2x-r/r(3)
From (2) & (3)
r/x−r=2x−r/r
⇒2x^2−3xr=0
x(2x−3r)=0
x=3r/2 (4)
From (1) & (4) we get,
xr=6
3 (r x r)/2=6
r^2=4
r=2
Hence (b) is the correct answer.

To find the radius of the circle inscribed in quadrilateral ABCD, we can use the formula for the inradius of a quadrilateral:

r = A / s

where r is the inradius, A is the area of the quadrilateral, and s is the semi-perimeter of the quadrilateral.

1. Identify the given information:
- Side AB is parallel to side CD.
- 2AB = CD.
- AD is perpendicular to AB and CD.
- The area of the quadrilateral is 18.

2. Determine the lengths of the sides:
Since AB is parallel to CD, we can consider AB and CD as the bases of the trapezoid ABCD, with AD as the height. Let AB = x and CD = 2x.

3. Calculate the area of the quadrilateral:
The area of a trapezoid is given by the formula:

A = (1/2) * (sum of the bases) * height

In this case, the area is given as 18, so we have:

18 = (1/2) * (x + 2x) * AD
18 = (3/2) * x * AD

4. Calculate the semi-perimeter:
The semi-perimeter of a quadrilateral is given by the formula:

s = (sum of the sides) / 2

In this case, the sum of the sides is:

AB + BC + CD + DA = x + AD + 2x + AD = 3x + 2AD

So, the semi-perimeter is:

s = (3x + 2AD) / 2

5. Substitute the values into the formula for the inradius:
Using the formula r = A / s, we can substitute the values of A and s:

r = 18 / [(3x + 2AD) / 2]
r = (36 / (3x + 2AD)

6. Simplify the expression:
To find the inradius, we need to eliminate the variable AD. We can use the Pythagorean theorem in triangle ADB to relate x and AD:

x^2 + AD^2 = (2x)^2
x^2 + AD^2 = 4x^2
AD^2 = 3x^2

Substituting AD^2 = 3x^2 into the expression for r:

r = (36 / (3x + 2AD)
r = (36 / (3x + 2sqrt(3x^2))
r = 36 / (3x + 2x(sqrt(3))
r = 36 / (5x(sqrt(3)))

7. Simplify further:
To simplify the expression, we can rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator:

r = (36 * (5x(sqrt(3)))) / ((5x(sqrt(3))) * (5x(sqrt(3))))
r = (180x(sqrt(3))) / (25x(sqrt(3))^2)
r = (180x(sqrt(3))) / (25x(3))
r = (180x(sqrt(3))) / (75x)
r = 2(sqrt(3))

Therefore, the radius of the circle inscribed in quadrilateral ABCD is 2(sqrt(3)), which corresponds to option B.