Printable math worksheet for calculating the area of mixed shapes, suitable for educational practice.

To solve the problem of finding the area of each figure in the worksheet, we need to use the appropriate formulas for the given shapes. Here are the steps and solutions for each figure:

Formulas Needed:

1. Circle: \( \text{Area} = \pi r^2 \)
2. Square: \( \text{Area} = s^2 \) (where \( s \) is the side length)
3. Rectangle: \( \text{Area} = l \times w \) (where \( l \) is the length and \( w \) is the width)
4. Triangle: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

Solutions:

#### 1. Circle (Radius = 5 ft)
\[ \text{Area} = \pi r^2 = \pi (5)^2 = 25\pi \]
\[ \text{Area} \approx 78.54 \, \text{ft}^2 \]

#### 2. Square (Side = 12 yd)
\[ \text{Area} = s^2 = (12)^2 = 144 \, \text{yd}^2 \]

#### 3. Square (Side = 6 in)
\[ \text{Area} = s^2 = (6)^2 = 36 \, \text{in}^2 \]

#### 4. Rectangle (Length = 7 in, Width = 3 m)
(Note: The units are inconsistent. Assuming the problem meant both dimensions in the same unit, let's assume it's all in inches for consistency.)
\[ \text{Area} = l \times w = 7 \times 3 = 21 \, \text{in}^2 \]

#### 5. Circle (Diameter = 4 ft)
\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{4}{2} = 2 \, \text{ft} \]
\[ \text{Area} = \pi r^2 = \pi (2)^2 = 4\pi \]
\[ \text{Area} \approx 12.57 \, \text{ft}^2 \]

#### 6. Triangle (Base = 8 yd, Height = 6 yd)
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24 \, \text{yd}^2 \]

#### 7. Triangle (Base = 1 yd, Height = 4 yd)
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 4 = 2 \, \text{yd}^2 \]

#### 8. Rectangle (Length = 10 in, Width = 2 in)
\[ \text{Area} = l \times w = 10 \times 2 = 20 \, \text{in}^2 \]

#### 9. Square (Side = 9 ft)
\[ \text{Area} = s^2 = (9)^2 = 81 \, \text{ft}^2 \]

#### 10. Rectangle (Length = 7 ft, Width = 6 ft)
\[ \text{Area} = l \times w = 7 \times 6 = 42 \, \text{ft}^2 \]

#### 11. Circle (Diameter = 10 yd)
\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{10}{2} = 5 \, \text{yd} \]
\[ \text{Area} = \pi r^2 = \pi (5)^2 = 25\pi \]
\[ \text{Area} \approx 78.54 \, \text{yd}^2 \]

#### 12. Triangle (Base = 15 in, Height = 4 in)
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times 4 = 30 \, \text{in}^2 \]

Final Answers:

1. \( 78.54 \, \text{ft}^2 \)
2. \( 144 \, \text{yd}^2 \)
3. \( 36 \, \text{in}^2 \)
4. \( 21 \, \text{in}^2 \)
5. \( 12.57 \, \text{ft}^2 \)
6. \( 24 \, \text{yd}^2 \)
7. \( 2 \, \text{yd}^2 \)
8. \( 20 \, \text{in}^2 \)
9. \( 81 \, \text{ft}^2 \)
10. \( 42 \, \text{ft}^2 \)
11. \( 78.54 \, \text{yd}^2 \)
12. \( 30 \, \text{in}^2 \)

Boxed Final Answer:

\[
\boxed{
\begin{array}{ll}
1. & 78.54 \, \text{ft}^2 \\
2. & 144 \, \text{yd}^2 \\
3. & 36 \, \text{in}^2 \\
4. & 21 \, \text{in}^2 \\
5. & 12.57 \, \text{ft}^2 \\
6. & 24 \, \text{yd}^2 \\
7. & 2 \, \text{yd}^2 \\
8. & 20 \, \text{in}^2 \\
9. & 81 \, \text{ft}^2 \\
10. & 42 \, \text{ft}^2 \\
11. & 78.54 \, \text{yd}^2 \\
12. & 30 \, \text{in}^2 \\
\end{array}
}
\]