Answer the following questions based on the information given below.
A circle with centre O and radius 2 cm inscribes a triangle ABC. Two tangents AD and BD are drawn from a point D outside the circle anf `angleABD=60^@`.
Find the area of `squareOADB`, in `cm^2`.

Options

A. `4sqrt(3)`
B. `3sqrt(3)`
C. 4
D. none of these

Answer

To find the area of the quadrilateral OADB, we will break down the problem step by step.

### Step 1: Understand the Geometry
We have a circle with center O and radius 2 cm. The tangents AD and BD are drawn from point D outside the circle, and we know that angle ABD = 60 degrees.

### Step 2: Identify Key Angles
Since AD and BD are tangents to the circle from point D, the angles between the radius and the tangent at points A and B are both 90 degrees. Therefore:
- Angle OAD = 90 degrees
- Angle OBD = 90 degrees

### Step 3: Calculate Angle AOB
Using the angle sum property in triangle ABD:
- Angle ADB = 180 - (Angle ABD + Angle BAD)
- Since AD = BD (tangents from the same point), Angle BAD = Angle ABD = 60 degrees.
- Therefore, Angle ADB = 180 - (60 + 60) = 60 degrees.

Now, we can find Angle AOB:
- Angle AOB = Angle AOD + Angle BOD = 60 + 60 = 120 degrees.

### Step 4: Use the Law of Cosines in Triangle AOB
We can use the Law of Cosines to find the length of AB:
- AB² = OA² + OB² - 2 * OA * OB * cos(Angle AOB)
- Since OA = OB = 2 cm (the radius), we have:
\[ AB² = 2² + 2² - 2 * 2 * 2 * cos(120^\circ) \]
- cos(120 degrees) = -1/2, so:
\[ AB² = 4 + 4 + 4 = 12 \implies AB = 2\sqrt{3} \text{ cm} \]

### Step 5: Calculate the Area of Triangle ADB
Since triangle ADB is isosceles with AB = 2√3 cm and angle ADB = 60 degrees, we can find the area using the formula:
\[ \text{Area} = \frac{1}{2} \times base \times height \]
To find the height, we can drop a perpendicular from D to AB, which bisects AB since triangle ADB is isosceles. The height can be calculated as:
- Height = \( AB \cdot \sin(60^\circ) = 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 3 \text{ cm} \)

Thus, the area of triangle ADB is:
\[ \text{Area}_{ADB} = \frac{1}{2} \times AB \times height = \frac{1}{2} \times 2\sqrt{3} \times 3 = 3\sqrt{3} \text{ cm}^2 \]

### Step 6: Calculate the Area of Triangle OAB
Now we calculate the area of triangle OAB:
\[ \text{Area}_{OAB} = \frac{1}{2} \times OA \times OB \times \sin(Angle AOB) \]
- OA = OB = 2 cm and Angle AOB = 120 degrees:
\[ \text{Area}_{OAB} = \frac{1}{2} \times 2 \times 2 \times \sin(120^\circ) = \frac{1}{2} \times 4 \times \frac{\sqrt{3}}{2} = \sqrt{3} \text{ cm}^2 \]

### Step 7: Calculate the Area of Quadrilateral OADB
Finally, we can find the area of quadrilateral OADB by adding the areas of triangles ADB and OAB:
\[ \text{Area}_{OADB} = \text{Area}_{ADB} + \text{Area}_{OAB} = 3\sqrt{3} + \sqrt{3} = 4\sqrt{3} \text{ cm}^2 \]

### Final Answer
The area of quadrilateral OADB is \( 4\sqrt{3} \text{ cm}^2 \).

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