Given:
- ABC is an equilateral triangle.
- AB, BC, and AC are tangents to the circle at points P, Q, and R respectively.
- The center of the circle is \( O \) and the radius is \( 8 \) cm.
Let's determine the length of PO, PC, PA, and the area of triangle ABC.
### Step 1: Understanding the problem
1. **Circle Tangents and Triangle Properties**:
- \( O \) is the center of the circle.
- The radius \( r = 8 \) cm.
- \( PA \), \( PB \), and \( PC \) touch the circle at points \( P \), \( Q \), and \( R \) respectively.
- Since \( ABC \) is an equilateral triangle, all sides are equal, and the distances from \( O \) to each tangent point (the radii) are all equal and perpendicular to the sides.
### Step 2: Calculate PO
Since \( O \) is the center, \( OP \) is one of the radii of the circle.
\[ PO = r = 8 \text{ cm} \]
### Step 3: Calculate PA
As all sides of the triangle are tangents to the circle, and \( ABC \) is an equilateral triangle, if we consider any of its vertex (say \( A \)):
- \( PA = PB = PC \)
Here the length \( PA \) from the vertex to the tangent point can be related to the side length \( AB \) of the equilateral triangle.
For tangents from point \( A \) to the circle,
\[ PA = \text{side length} \]
### Step 4: Find the side length of the equilateral triangle
Let the side length of \( \bigtriangleup ABC \) be \( s \).
The distance from the center \( O \) to any side (which is the altitude of \( \triangle OPB \), considering \( O \) to sideline \( BC \)):
\[ d = \frac{s \sqrt{3}}{6} \]
Since this is the distance from \( O \) to the midpoint of \( BC \) (altitude of the equilateral triangle to the side when the circle is inscribed):
\[ d = r \]
\[ \frac{s \sqrt{3}}{6} = 8 \]
\[ s \sqrt{3} = 48 \]
\[ s = \frac{48}{\sqrt{3}} \]
\[ s = \frac{48 \sqrt{3}}{3} \]
\[ s = 16\sqrt{3} \text{ cm} \]
### Step 5: Calculate PA, PB, PC as it is the side length
\[ PA = PB = PC = s = 16 \sqrt{3} \text{ cm} \]
### Step 6: Calculate the area of triangle ABC
For an equilateral triangle, the area \( A \) is given by:
\[ A = \frac{s^2 \sqrt{3}}{4} \]
Substitute \( s = 16 \sqrt{3} \):
\[ A = \frac{(16 \sqrt{3})^2 \sqrt{3}}{4} \]
\[ A = \frac{256 \times 3 \times \sqrt{3}}{4} \]
\[ A = \frac{768 \sqrt{3}}{4} \]
\[ A = 192 \sqrt{3} \text{ cm}^2 \]
### Summary
1. \( PO = 8 \text{ cm} \)
2. \( PA = PB = PC = 16 \sqrt{3} \text{ cm} \)
3. Area of \(\triangle ABC\) is \( 192 \sqrt{3} \text{ cm}^2 \)
ABC is an equilateral triangle. AB, BC and AC are tangents to the circle at the points P, Q, R. O is the center of the circle, radius 8 cm. Calculate the length of PO, PC, PA, Area of triangle ABC?
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