To solve for the area enclosed by the figure, let's break down the shape into two parts: a rectangle and a half-circle.

Step-by-Step Solution

1. Rectangle Area:

   • The dimensions of the rectangle are given as $6 \, \text{yd}$ (height) and $8 \, \text{yd}$ (width).
   • The area of a rectangle is calculated by: $\text{Area}_{\text{rectangle}} = \text{width} \times \text{height} = 8 \times 6 = 48 \, \text{yd}^2$

2. Half-Circle Area:

   • The diameter of the half-circle is equal to the width of the rectangle, which is $6 \, \text{yd}$.
   • Therefore, the radius $r$ is: $r = \frac{6}{2} = 3 \, \text{yd}$
   • The area of a full circle is given by $\pi r^2$. For a half-circle, we take half of this area: $\text{Area}_{\text{half-circle}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.14 \times 3^2$ $= \frac{1}{2} \times 3.14 \times 9 = \frac{1}{2} \times 28.26 = 14.13 \, \text{yd}^2$

3. Total Area:

   • Adding the areas of the rectangle and half-circle together: $\text{Total Area} = 48 + 14.13 = 62.13 \, \text{yd}^2$

Final Answer

The area enclosed by the figure is: $62.13 \, \text{yd}^2$

Would you like further explanation on any of the steps?

Related Questions

1. What would the area be if the entire circle was included instead of a half-circle?
2. How does changing the diameter of the half-circle affect the total area?
3. If the width of the rectangle changed to 10 yards, how would that change the area?
4. What is the perimeter of the figure, including the straight edges and the curved edge?
5. How can you use integration to calculate the area of more complex shapes?

Tip:

When working with composite shapes, break them down into simpler shapes (like rectangles, triangles, and circles) to find the area easily.