To find the base side length of the triangular pyramid, we first need to calculate the volume of the pyramid using the given mass and density. The formula to calculate volume from mass and density is:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Given:
- Mass \( = 400 \) grams
- Density \( = 16 \) g/cm³
Substituting the values:
\[ \text{Volume} = \frac{400 , \text{g}}{16 , \text{g/cm}^3} = 25 , \text{cm}^3 \]
Next, we use the formula for the volume of a pyramid:
\[ \text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \]
We know the volume (25 cm³) and the height (5 cm):
\[ 25 = \frac{1}{3} \times \text{Base Area} \times 5 \]
Rearranging for Base Area:
\[ \text{Base Area} = \frac{25 \times 3}{5} = 15 , \text{cm}^2 \]
Since the base is an equilateral triangle, we can find the area using the formula for the area of an equilateral triangle:
\[ \text{Area} = \frac{\sqrt{3}}{4} s^2 \]
where \( s \) is the length of a side of the triangle. Setting this equal to the base area we calculated:
\[ \frac{\sqrt{3}}{4} s^2 = 15 \]
Solving for \( s^2 \):
\[ s^2 = \frac{15 \times 4}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]
Taking the square root to find \( s \):
\[ s = \sqrt{20\sqrt{3}} = \sqrt{20} \cdot \sqrt[4]{3} = 2\sqrt{5} \cdot 3^{0.25} \]
To get an approximate numerical value, calculate \( \sqrt{20} \approx 4.47 \) and \( \sqrt[4]{3} \approx 1.316 \):
\[ s \approx 4.47 \cdot 1.316 \approx 5.89 , \text{cm} \]
Therefore, the correct answer is \( 5.89 , \text{cm} \).