To find the surface area of a pyramid with a square base, we need to calculate both the area of the base and the area of the triangular faces.

The surface area $A$ of the pyramid is given by:

$A = A_{\text{base}} + A_{\text{lateral faces}}$

Step 1: Area of the base

The base is a square, and the side length of the square is $m = 5$ inches. The area of the square base is:

$A_{\text{base}} = m^2 = 5^2 = 25 \, \text{in}^2$

Step 2: Area of the lateral faces

The lateral faces of the pyramid are triangles. Each triangle has a base equal to $m = 5$ inches, and the slant height of each triangle is denoted by $e = 5$ inches.

The area of one triangular face is:

$A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 5 \times 5 = 12.5 \, \text{in}^2$

There are four triangular faces, so the total area of the lateral faces is:

$A_{\text{lateral faces}} = 4 \times 12.5 = 50 \, \text{in}^2$

Step 3: Total surface area

Now, we can find the total surface area by adding the area of the base and the lateral faces:

$A = A_{\text{base}} + A_{\text{lateral faces}} = 25 + 50 = 75 \, \text{in}^2$

Thus, the total surface area of the pyramid is 75 in², but none of the options match exactly. Let's check the information in the problem, where the slant height $e = 5$ is used.

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