The tangents drawn at point A and B of a circle with centre O, meet at P. If `angle AOB = 120^(@)` and AP = 6 cm, then what is the area of triangle APB (in `cm^(2)`) ?

Options

A. `6sqrt(3)`
B. `8sqrt(3)`
C. 9
D. `9sqrt(3)`

Answer

To find the area of triangle APB given the conditions in the problem, we can follow these steps:

### Step 1: Understand the Geometry
We have a circle with center O, and tangents AP and BP meeting at point P. The angle AOB is given as 120 degrees. Since AP and BP are tangents to the circle, we know that OA and OB are radii of the circle.

### Step 2: Identify Angles
Since OA and OB are radii, the angles OAP and OBP are both right angles (90 degrees). Therefore, we can find angle APB using the fact that the angles in triangle AOB sum up to 180 degrees:
\[ \angle APB = 180^\circ - \angle AOB - \angle OAP - \angle OBP = 180^\circ - 120^\circ - 90^\circ - 90^\circ = 60^\circ \]

### Step 3: Use Properties of Isosceles Triangle
Since AP = PB (both are tangents from the same external point P), triangle APB is isosceles with AP = PB = 6 cm. We already found that angle APB = 60 degrees.

### Step 4: Calculate Area of Triangle APB
The area \( A \) of triangle APB can be calculated using the formula for the area of a triangle:
\[ A = \frac{1}{2} \times base \times height \]
In triangle APB, we can use the formula for the area of an isosceles triangle:
\[ A = \frac{1}{2} \times AP \times PB \times \sin(\angle APB) \]
Substituting the known values:
\[ A = \frac{1}{2} \times 6 \times 6 \times \sin(60^\circ) \]
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ A = \frac{1}{2} \times 6 \times 6 \times \frac{\sqrt{3}}{2} = \frac{36\sqrt{3}}{4} = 9\sqrt{3} \text{ cm}^2 \]

### Final Answer
The area of triangle APB is \( 9\sqrt{3} \text{ cm}^2 \).

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