To find the base side length of the triangular pyramid, we can use the formula for the volume of a triangular pyramid (a tetrahedron):
\[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \]
We also know that the mass of the pyramid and the density can be used to find the volume:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \quad \Rightarrow \quad \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Given:
- Mass = 400 grams
- Density = 16 g/cm³
First, we calculate the volume:
\[ \text{Volume} = \frac{400 \text{ g}}{16 \text{ g/cm}^3} = 25 \text{ cm}^3 \]
Now we can set the volume equal to the pyramid volume formula:
\[ 25 = \frac{1}{3} \times \text{Base Area} \times 5 \]
Rearranging gives:
\[ \text{Base Area} = \frac{25 \times 3}{5} = 15 \text{ cm}^2 \]
The base of the pyramid is an equilateral triangle. The area \( A \) of an equilateral triangle with side length \( s \) is given by:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
Setting this equal to the base area we calculated, we have:
\[ \frac{\sqrt{3}}{4} s^2 = 15 \]
Now, solve for \( s^2 \):
\[ s^2 = \frac{15 \times 4}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]
Taking the square root of both sides to find \( s \):
\[ s = \sqrt{20\sqrt{3}} = \sqrt{20} \cdot \sqrt[4]{3} = \sqrt{20} \cdot 3^{1/4} \]
Calculating \( \sqrt{20} \):
\[ \sqrt{20} = 2\sqrt{5} \approx 2 \cdot 2.236 = 4.472 \]
The fourth root of 3 is approximately \( 1.316 \), so:
\[ s \approx 4.472 \cdot 1.316 \approx 5.89 \text{ cm} \]
Thus, the correctly calculated base side length of the triangular pyramid is:
5.89 cm.