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The problem involves a triangle \( ABC \) with a shaded region \( ADD \), which is part of a circle centered at \( A \). Here's a breakdown of the question and how we can approach it:

### Given:
- Triangle \( ABC \) is isosceles with \( AB = AC = r \) and angle \( \angle BAC = \theta \) radians.
- Point \( D \) lies on arc \( BC \) of the circle with center \( A \), making \( ADD \) the sector of the circle.

#### Part (a): Express the area of the shaded region in terms of \( r \) and \( \theta \).

The shaded region consists of the area of triangle \( ABC \) subtracted from the area of sector \( ADD \).

- **Area of sector \( ADD \)**: The area of a sector of a circle is given by:
  \[
  \text{Area of sector} = \frac{1}{2} r^2 \theta
  \]
  where \( r \) is the radius and \( \theta \) is the central angle in radians.

- **Area of triangle \( ABC \)**: For an isosceles triangle with two equal sides \( r \) and angle \( \theta \) between them, the area can be calculated using:
  \[
  \text{Area of triangle} = \frac{1}{2} r^2 \sin(\theta)
  \]

Thus, the area of the shaded region is:
\[
\text{Shaded area} = \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin(\theta)
\]
This simplifies to:
\[
\text{Shaded area} = \frac{1}{2} r^2 (\theta - \sin(\theta))
\]

#### Part (b): When \( r = 10 \) and \( \theta = \frac{\pi}{3} \), find the perimeter of the shaded region.

The perimeter of the shaded region consists of two parts:
1. The arc length from \( B \) to \( D \) (which is part of the sector).
2. The straight line \( BD \).

- **Arc length \( BD \)**: The arc length is given by the formula:
  \[
  \text{Arc length} = r \theta = 10 \times \frac{\pi}{3} = \frac{10\pi}{3}
  \]

- **Length of chord \( BD \)**: Using the formula for the length of a chord in a circle:
  \[
  BD = 2r \sin\left(\frac{\theta}{2}\right)
  \]
  Substituting \( r = 10 \) and \( \theta = \frac{\pi}{3} \):
  \[
  BD = 2 \times 10 \times \sin\left(\frac{\pi}{6}\right) = 20 \times \frac{1}{2} = 10
  \]

Thus, the total perimeter of the shaded region is:
\[
\text{Perimeter} = \text{Arc length} + \text{Chord length} = \frac{10\pi}{3} + 10
\]

### Final Answer:
- Part (a): The area of the shaded region is \( \frac{1}{2} r^2 (\theta - \sin(\theta)) \).
- Part (b): The perimeter of the shaded region when \( r = 10 \) and \( \theta = \frac{\pi}{3} \) is \( \frac{10\pi}{3} + 10 \).

Would you like further details or have any questions?

### Related Questions:
1. What would be the area of the triangle \( ABC \) when \( \theta = \frac{\pi}{4} \)?
2. How do you derive the formula for the arc length of a sector in a circle?
3. What happens to the area of the shaded region as \( \theta \) approaches \( 2\pi \)?
4. Can you compute the area of the full circle if \( \theta = 2\pi \) and \( r = 5 \)?
5. How do we find the chord length when the angle \( \theta \) is not given directly?

### Tip:
Remember that in any sector or circle-based problem, always make sure to express angles in radians, not degrees, to use standard formulas directly.

In the diagram, ABC is an isosceles triangle with AB = AC = r and angle BAC = θ radians. The point D lies on arc BC and ∠ADD is a sector of a circle with center A.

(a) Express the area of the shaded region in terms of r and θ.
(b) In the case where r = 10 and θ = π/3, find the perimeter of the shaded region.

Explore how to calculate the area of a shaded region within a circle and an isosceles triangle, using trigonometry and geometry concepts. Suitable for high school students learning about circular sectors and trigonometric formulas.

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