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### 1. **Determine the perimeter and area of the following figures:**

#### a) **Circle with radius 3 inches**
- **Perimeter (Circumference):** The formula for the circumference of a circle is:
  \[
  C = 2\pi r
  \]
  where \( r = 3 \) inches. Substituting:
  \[
  C = 2\pi(3) = 6\pi \approx 18.85 \text{ inches}
  \]

- **Area:** The formula for the area of a circle is:
  \[
  A = \pi r^2
  \]
  Substituting \( r = 3 \) inches:
  \[
  A = \pi(3)^2 = 9\pi \approx 28.27 \text{ square inches}
  \]

#### b) **Rectangle with dimensions 6 cm by 2 cm**
- **Perimeter:** The perimeter of a rectangle is:
  \[
  P = 2(l + w)
  \]
  where \( l = 6 \) cm and \( w = 2 \) cm:
  \[
  P = 2(6 + 2) = 2(8) = 16 \text{ cm}
  \]

- **Area:** The area of a rectangle is:
  \[
  A = l \times w = 6 \times 2 = 12 \text{ square cm}
  \]

#### c) **Right triangle with sides 8 cm, 9 cm, and 10 cm**
- **Perimeter:** The perimeter of a triangle is the sum of all its sides:
  \[
  P = 8 + 9 + 10 = 27 \text{ cm}
  \]

- **Area:** The area of a right triangle is:
  \[
  A = \frac{1}{2} \times \text{base} \times \text{height}
  \]
  Using the base as 8 cm and the height as 9 cm:
  \[
  A = \frac{1}{2} \times 8 \times 9 = 36 \text{ square cm}
  \]

#### d) **Irregular figure (L-shaped)**
- **Perimeter:** Sum the lengths of all sides:
  \[
  P = 4 + 7 + 4 + 3 + 6 + 3 + 7 + 4 + 6 = 44 \text{ cm}
  \]

- **Area:** Divide the shape into smaller rectangles. Each of the two main sections is:
  - Rectangle 1: \( 4 \text{ cm} \times 7 \text{ cm} = 28 \text{ square cm} \)
  - Rectangle 2: \( 6 \text{ cm} \times 3 \text{ cm} = 18 \text{ square cm} \)
  \[
  \text{Total area} = 28 + 18 = 46 \text{ square cm}
  \]

### 2. **Paint the lines of a basketball key:**

The basketball key has two components:
1. A rectangle with dimensions 10 ft by 14 ft.
2. A semicircle with a radius of 6 ft.

- **Perimeter of the rectangle (except the base):**
  \[
  P_{\text{rectangle}} = 14 + 10 + 14 = 38 \text{ ft}
  \]

- **Perimeter of the semicircle:**
  The formula for the perimeter of a semicircle is:
  \[
  P_{\text{semicircle}} = \pi r
  \]
  where \( r = 6 \) ft:
  \[
  P_{\text{semicircle}} = \pi(6) \approx 18.85 \text{ ft}
  \]

- **Total perimeter to paint:**
  \[
  P_{\text{total}} = 38 + 18.85 = 56.85 \text{ ft}
  \]

Since each can of paint covers 10 ft, the number of cans required is:
\[
\frac{56.85}{10} \approx 5.69
\]
Thus, you will need 6 cans of paint.

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Would you like further explanations on any of these solutions?

Here are some related questions you could explore:
1. How do you calculate the area of a sector in a circle?
2. How does the Pythagorean theorem apply to right triangles?
3. What is the difference between area and perimeter?
4. How would the solution change for a semicircle if the diameter were used instead of the radius?
5. What other shapes could you divide the irregular figure into to make the area calculation easier?

**Tip:** For complex shapes, dividing them into simple parts like rectangles or triangles helps make calculations easier.

Determine the perimeter and area of various geometric shapes and calculate how many cans of paint are needed to cover the perimeter of a basketball key.

Learn how to calculate the perimeter and area of various geometric shapes, including circles, rectangles, triangles, and irregular L-shaped figures. Additionally, calculate the number of paint cans required to cover the perimeter of a basketball key with a semicircle. Suitable for students in Grades 8-10.

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