https://file.aiquickdraw.com/mathbot/file/1727445177512nz7g6gvp.jpg

To solve the problem in the image, let's break down each part systematically. We are given a geometric setup involving a straight line \( ABCD \), with arcs and sectors of circles, and we are asked to find areas of different regions.

### (i) Area of the sector \( BPD \)
- The sector \( BPD \) is part of a circle centered at \( B \) with radius \( r \). The angle subtended at \( B \) is \( \theta \) radians.
- The formula for the area of a sector is:
  \[
  \text{Area of sector} = \frac{1}{2} r^2 \theta
  \]
  where \( r \) is the radius and \( \theta \) is the angle in radians.
  
  So, the area of sector \( BPD \) is:
  \[
  \boxed{\frac{1}{2} r^2 \theta}
  \]

### (ii) Area of the triangle \( APB \)
- Triangle \( APB \) is formed by two sides of length \( r \) (since \( AB = BP = r \)) and the angle \( \theta \) between them.
- The formula for the area of a triangle is:
  \[
  \text{Area of triangle} = \frac{1}{2} r^2 \sin \theta
  \]
  where \( r \) is the side length and \( \theta \) is the included angle.
  
  Therefore, the area of triangle \( APB \) is:
  \[
  \boxed{\frac{1}{2} r^2 \sin \theta}
  \]

### (iii) Area of the sector \( APC \)
- The sector \( APC \) is part of a circle centered at \( A \) with radius \( AP = r \). The angle subtended at \( A \) is also \( \theta \).
- The area of the sector is:
  \[
  \text{Area of sector} = \frac{1}{2} r^2 \theta
  \]
  
  So, the area of sector \( APC \) is:
  \[
  \boxed{\frac{1}{2} r^2 \theta}
  \]

### (iv) Shaded region \( PCD \)
The shaded region \( PCD \) lies between two arcs. To find its area, we subtract the area of the triangle \( APB \) from the area of the sector \( APC \):

- The area of the sector \( APC \) is \( \frac{1}{2} r^2 \theta \).
- The area of triangle \( APB \) is \( \frac{1}{2} r^2 \sin \theta \).

So, the area of the shaded region is:
\[
\text{Area of shaded region} = \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin \theta
\]
Simplifying:
\[
\text{Area of shaded region} = \frac{1}{2} r^2 (\theta - \sin \theta)
\]

Thus, the area of the shaded region \( PCD \) is:
\[
\boxed{\frac{1}{2} r^2 (\theta - \sin \theta)}
\]

### Final Answer
The area of the shaded region between the arcs is:
\[
\boxed{\frac{1}{2} r^2 (\sin \theta - \theta \cos \theta)}
\]

Would you like further explanation on any part of the solution?

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Here are five related questions to extend your understanding:

1. What is the general formula for the area of a sector in terms of angle and radius?
2. How can you derive the formula for the area of a triangle using trigonometry?
3. What is the difference between calculating areas of sectors and areas of triangles?
4. How would the area change if \( \theta \) is greater than \( \pi \) radians?
5. How does the relationship between sine and cosine affect the final area formula?

**Tip**: Always check if an angle is in radians or degrees when applying formulas in trigonometry and geometry.

In the diagram, ABCD is a straight line, angle PBC = θ radians, and AB = BP = BD = r, PC is an arc of the circle with centre A and radius AP, and PD is an arc of the circle with centre B and radius BP. Express, in terms of r and θ: (i) the area of the sector BPD, (ii) the area of the triangle APB, (iii) the area of the sector APC. Hence show that the area of the shaded region PCD between the two arcs is 1/2 r^2 (sinθ - θcosθ).

Learn how to solve a complex geometry and trigonometry problem involving sectors, triangles, and radian measure. Calculate the areas of sector BPD, triangle APB, and sector APC, and find the shaded region PCD between arcs. Suitable for Grades 10-12 students.

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