To find the base side length of the triangular pyramid, we can use the formula for the volume of a pyramid and relate it to its mass and density.
First, we have the mass \( m \) of the pyramid given as 400 grams and the density \( \rho \) as 16 g/cm³. The volume \( V \) can be calculated using the formula:
\[ V = \frac{m}{\rho} \]
Substituting the values for \( m \) and \( \rho \):
\[ V = \frac{400 \text{ grams}}{16 \text{ g/cm}^3} = 25 \text{ cm}^3 \]
Next, the volume \( V \) of a pyramid is also given by the formula:
\[ V = \frac{1}{3} \cdot B \cdot h \]
where \( B \) is the area of the base and \( h \) is the height. Given that the height \( h \) is 5 cm, we can set up our equation:
\[ 25 = \frac{1}{3} \cdot B \cdot 5 \]
Solving for \( B \):
\[ 25 = \frac{5}{3} \cdot B \]
\[ B = 25 \cdot \frac{3}{5} = 15 \text{ cm}^2 \]
Now, since the base of the pyramid is an equilateral triangle, we can calculate the area \( B \) using the formula for the area of an equilateral triangle:
\[ B = \frac{\sqrt{3}}{4} s^2 \]
where \( s \) is the side length. Setting this equal to our calculated area:
\[ \frac{\sqrt{3}}{4} s^2 = 15 \]
Solving for \( s^2 \):
\[ s^2 = 15 \cdot \frac{4}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]
Now taking the square root of both sides:
\[ s = \sqrt{20\sqrt{3}} = \sqrt{20 \cdot \sqrt{3}} = \sqrt{20} \cdot \sqrt[4]{3} \]
Calculating \( \sqrt{20} \):
\[ \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \]
So we have:
\[ s = 2 \sqrt{5 \sqrt{3}} = 2 (\text{approximately } 5.89) \]
Thus, the possible answer choices presented seem to include 5.89 cm as the correct base side length of the triangular pyramid.
Therefore, the correct answer is:
5.89 cm