A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the `Delta ABC`

Options

A. 12 cm
B. 20 cm
C. 24 cm
D. 30 cm

Explanation

To find the perimeter of triangle ABC formed by the tangents from point A to the circle and the tangent BC at point R on the minor arc PQ, we can follow these steps:

### Step 1: Understand the Geometry
We have a circle with center O and radius 5 cm. Point A is located 13 cm from O. Tangents AP and AQ are drawn from point A to points P and Q on the circle. A tangent BC is drawn at point R, which lies on the minor arc PQ.

### Step 2: Calculate the Length of Tangents AP and AQ
Using the Pythagorean theorem in triangle OAP:
- OA = 13 cm (distance from center O to point A)
- OP = 5 cm (radius of the circle)

According to the Pythagorean theorem:
\[ OA^2 = AP^2 + OP^2 \]
Substituting the known values:
\[ 13^2 = AP^2 + 5^2 \]
\[ 169 = AP^2 + 25 \]
\[ AP^2 = 169 - 25 \]
\[ AP^2 = 144 \]
Taking the square root:
\[ AP = \sqrt{144} = 12 \text{ cm} \]

Since AP = AQ (tangents from a point outside the circle are equal), we have:
\[ AQ = 12 \text{ cm} \]

### Step 3: Use the Properties of Tangents
Since BC is a tangent at point R, we can use the property of tangents from an external point:
- BR = BP (tangent from B to the circle)
- CR = CQ (tangent from C to the circle)

### Step 4: Express the Perimeter of Triangle ABC
The perimeter of triangle ABC is given by:
\[ \text{Perimeter} = AB + BC + CA \]

Since BC can be expressed as:
\[ BC = BR + RC \]
We can substitute:
\[ \text{Perimeter} = AB + (BR + RC) + CA \]

Using the properties of tangents:
- BR = BP
- RC = CQ

Thus:
\[ \text{Perimeter} = AB + BP + CQ + CA \]

### Step 5: Substitute Known Values
Since we know:
- AP = AQ = 12 cm
- AB = AP = 12 cm
- AC = AQ = 12 cm

We can substitute:
\[ \text{Perimeter} = AB + BP + CQ + CA \]
\[ \text{Perimeter} = AP + CQ + AP \]
\[ \text{Perimeter} = 12 + 12 + 12 = 24 \text{ cm} \]

### Conclusion
The perimeter of triangle ABC is:
\[ \text{Perimeter} = 24 \text{ cm} \]

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