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### Problem 4(b)
- **Diagram**: A square inscribed in a circle.
- **Given**: The radius of the circle is \( r = 12 \) cm.
- **Objective**: Find the side of the square.

**Solution**:
The diagonal of the square is equal to the diameter of the circle.

1. Diameter of the circle = \( 2r = 2 \times 12 = 24 \) cm.
2. Let the side of the square be \( s \). The diagonal of the square is given by \( s\sqrt{2} \).
3. Equating the diagonal to the diameter of the circle:
   \[
   s\sqrt{2} = 24
   \]
   \[
   s = \frac{24}{\sqrt{2}} = \frac{24 \times \sqrt{2}}{2} = 12\sqrt{2} \text{ cm}
   \]

### Problem 4(c)
- **Diagram**: A circle inscribed in a square.
- **Given**: Side of the square is \( s = 28 \) cm.
- **Objective**: Find the radius of the circle.

**Solution**:
The diameter of the circle is equal to the side of the square.

1. Diameter of the circle = \( 28 \) cm.
2. Radius \( r \) of the circle is:
   \[
   r = \frac{28}{2} = 14 \text{ cm}
   \]

### Problem 4(d)
- **Diagram**: A sector of a circle.
- **Given**: The radius is \( r = 7 \) cm.
- **Objective**: Find the area of the sector (if the angle is provided).

**Solution**:
The area of the sector is given by:
\[
\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2
\]
where \( \theta \) is the central angle in degrees. If \( \theta \) isn't provided, we cannot calculate the exact area. Assuming \( \theta \) was provided in the problem:

If \( \theta = 60^\circ \):
\[
\text{Area} = \frac{60^\circ}{360^\circ} \times \pi \times 7^2 = \frac{1}{6} \times \pi \times 49 = \frac{49\pi}{6} \text{ square cm}
\]

### Problem 4(e)
- **Diagram**: A circle within a square with the side of the square \( s = 15 \) cm.
- **Objective**: Find the area of the shaded region outside the circle but inside the square.

**Solution**:
1. Area of the square = \( s^2 = 15^2 = 225 \text{ square cm} \).
2. Diameter of the circle = \( s = 15 \) cm, hence radius \( r = 7.5 \) cm.
3. Area of the circle = \( \pi r^2 = \pi \times 7.5^2 = 56.25\pi \text{ square cm} \).
4. Area of the shaded region = Area of the square - Area of the circle:
   \[
   \text{Shaded Area} = 225 - 56.25\pi \text{ square cm}
   \]

### Problem 4(f)
- **Diagram**: A semicircle on top of a rectangle.
- **Given**: Diameter of the semicircle is equal to the length of the rectangle, which is 28 cm.
- **Objective**: Find the perimeter of the figure.

**Solution**:
1. Length of the rectangle = 28 cm, hence the radius of the semicircle \( r = \frac{28}{2} = 14 \) cm.
2. Perimeter of the figure = Perimeter of rectangle (excluding the length on which semicircle is placed) + Circumference of semicircle:
   \[
   \text{Perimeter} = 2b + 2r + \pi r = 2b + r(\pi + 2)
   \]
   where \( b \) is the width of the rectangle (not provided).

   The perimeter formula involves adding the sides of the rectangle and the curve of the semicircle. Without the width, the perimeter cannot be determined precisely, but the formula above gives the method.

Would you like a detailed explanation of any part, or do you have any specific questions? 

### Related Questions:
1. How is the diagonal of a square related to the side length?
2. What is the relationship between the area and the perimeter of a circle?
3. How do you calculate the area of a sector if the angle is provided?
4. How would the shaded area change if the circle in the square were smaller?
5. What would the formula for the perimeter of a semicircle look like if the rectangle's width were known?

### Tip:
When dealing with circles, always remember the fundamental relationships: \( C = 2\pi r \) for the circumference and \( A = \pi r^2 \) for the area.

Explore solutions to geometry problems involving inscribed shapes, circle properties, and more. Includes detailed steps and formulas for finding areas, radii, and perimeters. Suitable for Grades 9-12.

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