From a point A which is at a distance of 10 cm from the centre O of radius 6 cm, the pair of tangents AB and AC to the circle are drawn. What will be the area of the quadrilateral ABOC?

Answer

To find the area of the quadrilateral ABOC, we can follow these steps:

### Step 1: Understand the configuration
We have a circle with center O and radius 6 cm. Point A is located 10 cm away from the center O. Tangents AB and AC are drawn from point A to the circle.

### Step 2: Identify the right triangles
Since the radius is perpendicular to the tangent at the point of tangency, triangles OAB and OAC are right triangles.

### Step 3: Use the Pythagorean theorem
In triangle OAC:
- OA = 10 cm (distance from A to O)
- OC = 6 cm (radius of the circle)
- AC is the tangent we need to find.

Using the Pythagorean theorem:
\[ OA^2 = OC^2 + AC^2 \]
\[ 10^2 = 6^2 + AC^2 \]
\[ 100 = 36 + AC^2 \]
\[ AC^2 = 100 - 36 \]
\[ AC^2 = 64 \]
\[ AC = \sqrt{64} = 8 \text{ cm} \]

### Step 4: Calculate the area of triangle OAC
The area of triangle OAC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, OC is the base and AC is the height.
\[ \text{Area of } OAC = \frac{1}{2} \times OC \times AC \]
\[ = \frac{1}{2} \times 6 \times 8 \]
\[ = \frac{1}{2} \times 48 = 24 \text{ cm}^2 \]

### Step 5: Find the area of quadrilateral ABOC
Since triangles OAC and OAB are congruent (by SSS rule), the area of triangle OAB is also 24 cm².
Thus, the area of quadrilateral ABOC is:
\[ \text{Area of } ABOC = \text{Area of } OAC + \text{Area of } OAB \]
\[ = 24 + 24 = 48 \text{ cm}^2 \]

### Final Answer
The area of quadrilateral ABOC is **48 cm²**.

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