From a point `P` which is at a distance of `13` `cm` from the center `O` of a circle of radius `5` `cm`, the pair of tangents `PQ` and `PR` to the circle is drawn. Then, the area of the quadrilateral `PQOR` is

Options

A. `60cm^(2)`
B. `65cm^(2)`
C. `30cm^(2)`
D. `32.5cm^(2)`

Explanation

To solve the problem of finding the area of quadrilateral \(PQOR\), we will follow these steps:

### Step 1: Understand the Configuration
We have a circle with center \(O\) and radius \(5 \, \text{cm}\). A point \(P\) is located \(13 \, \text{cm}\) away from the center \(O\). From point \(P\), two tangents \(PQ\) and \(PR\) are drawn to the circle.

### Step 2: Identify Right Triangles
Since the tangents from a point outside the circle are perpendicular to the radius at the point of contact, we know that:
- \( \angle PQO = 90^\circ \)
- \( \angle PRO = 90^\circ \)

This means that triangles \(PQO\) and \(PRO\) are right triangles.

### Step 3: Apply the Pythagorean Theorem
In triangle \(PQO\), we can apply the Pythagorean theorem:
\[ OP^2 = OQ^2 + PQ^2 \]
Where:
- \(OP = 13 \, \text{cm}\) (distance from point \(P\) to center \(O\))
- \(OQ = 5 \, \text{cm}\) (radius of the circle)
- \(PQ\) is the length of the tangent we want to find.

Substituting the known values:
\[ 13^2 = 5^2 + PQ^2 \]
\[ 169 = 25 + PQ^2 \]
\[ PQ^2 = 169 - 25 = 144 \]
\[ PQ = \sqrt{144} = 12 \, \text{cm} \]

### Step 4: Calculate the Area of Triangle \(PQO\)
The area \(A\) of triangle \(PQO\) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \(OQ = 5 \, \text{cm}\) and the height \(PQ = 12 \, \text{cm}\):
\[ A_{PQO} = \frac{1}{2} \times 5 \times 12 = \frac{60}{2} = 30 \, \text{cm}^2 \]

### Step 5: Calculate the Area of Triangle \(POR\)
Since triangle \(POR\) is congruent to triangle \(PQO\) (both have the same base and height):
\[ A_{POR} = 30 \, \text{cm}^2 \]

### Step 6: Calculate the Area of Quadrilateral \(PQOR\)
The area of quadrilateral \(PQOR\) is the sum of the areas of triangles \(PQO\) and \(POR\):
\[ A_{PQOR} = A_{PQO} + A_{POR} = 30 + 30 = 60 \, \text{cm}^2 \]

Thus, the area of quadrilateral \(PQOR\) is \(60 \, \text{cm}^2\).

### Final Answer
The area of quadrilateral \(PQOR\) is \( \boxed{60 \, \text{cm}^2} \).

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