To find out how many square inches of cardboard Leonard will need for his pyramid, we can use the formula provided:
\[ \text{Surface Area} = 4\left(\frac{1}{2}bh\right) + b_2 \]
Where:
- \( b \) is the base length.
- \( h \) is the height.
- \( b_2 \) is the area of the base.
For a square pyramid:
- The base length \( b \) is 3 inches.
- The height \( h \) is 8 inches.
First, let's calculate the area of the base:
\[ b_2 = \text{base length}^2 = 3^2 = 9 \text{ square inches} \]
Next, we calculate the area of the triangular faces of the pyramid. Each triangular face has a base \( b \) and height which is the slant height of the triangle. However, since we only have the height of the pyramid (vertical height) given, we cannot calculate the slant height directly from the given data.
Typically, for the pyramid's triangular sides, if the base (the base of the triangle) of the triangular face is the same as the base of the pyramid. Thus, the area of one triangular face is:
\[ \text{Area of one triangular face} = \frac{1}{2} \times \text{base} \times \text{height} \]
Since there are 4 triangular faces, we multiply by 4:
Assuming h = 8 inches is the vertical height, and the triangular height can be assumed based on a simple triangular face calculation.
However, to focus only on the data we can manipulate without making unconfirmed assumptions here. The surface area of triangular faces must base its total height on the pyramid geometry and scale directly to h as it relates to the surface context.
For simplicity:
Using the height \( h = 8 \) gives layers into the triangular conceptualization from a linear ratio perspective, the same base gives:
\[ \text{Area of 4 triangular faces} = 4\left(\frac{1}{2} \times 3 \times h\right) = 4\left(\frac{1}{2} \times 3 \times 8\right) = 4 \times 12 = 48 \text{ square inches} \]
Adding the base area:
\[ \text{Total Surface Area} = 48 + 9 = 57 \text{ square inches} \]
Thus, Leonard will need a total of 57 square inches of cardboard for his pyramid.