A pyramid 8cm high stands on a square base 6cm calculate the total surface area

The height of the pyramid ($h$) is 8 cm.
The side length of the square base ($s$) is 6 cm.

Area of base = $s^2$
Area of base = $6^2$ cm$^2$
Area of base = 36 cm$^2$

The slant height can be found using the Pythagorean theorem, considering the height of the pyramid, half the base side length, and the slant height as a right triangle.
$l = \sqrt{h^2 + (\frac{s}{2})^2}$
$l = \sqrt{8^2 + (\frac{6}{2})^2}$
$l = \sqrt{8^2 + 3^2}$
$l = \sqrt{64 + 9}$
$l = \sqrt{73}$ cm

The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. In this case, the base of the triangle is the side of the square base ($s$), and the height of the triangle is the slant height ($l$).
Area of one triangular face = $\frac{1}{2} \times s \times l$
Area of one triangular face = $\frac{1}{2} \times 6 \times \sqrt{73}$ cm$^2$
Area of one triangular face = $3\sqrt{73}$ cm$^2$

Since the base is square, there are four identical triangular faces.
Total area of triangular faces = $4 \times (\text{Area of one triangular face})$
Total area of triangular faces = $4 \times 3\sqrt{73}$ cm$^2$
Total area of triangular faces = $12\sqrt{73}$ cm$^2$

Total surface area = Area of base + Total area of triangular faces
Total surface area = $36 + 12\sqrt{73}$ cm$^2$

$\sqrt{73} \approx 8.544$
Total surface area $\approx 36 + 12 \times 8.544$
Total surface area $\approx 36 + 102.528$
Total surface area $\approx 138.528$ cm$^2$

The total surface area of the pyramid is $36 + 12\sqrt{73}$ cm$^2$.