To determine the base side length of the triangular pyramid, we'll first use the relationship between mass, density, and volume to find the volume of the pyramid. The formula that relates these variables is:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Given:
- Mass \(m = 400 , \text{grams}\)
- Density \(\rho = 16 , \text{g/cm}^3\)
Let's calculate the volume of the pyramid:
\[ \text{Volume} = \frac{400 , \text{grams}}{16 , \text{g/cm}^3} = 25 , \text{cm}^3 \]
Next, we know the formula for the volume \(V\) of a triangular pyramid (also known as a tetrahedron) is given by:
\[ V = \frac{1}{3} \times B \times h \]
Where:
- \(B\) is the area of the base,
- \(h\) is the height of the pyramid.
Given that the height \(h\) is 5 cm, we rearrange the equation to solve for the base area \(B\):
\[ B = \frac{3V}{h} \]
Substituting in the known values:
\[ B = \frac{3 \times 25 , \text{cm}^3}{5 , \text{cm}} = \frac{75 , \text{cm}^3}{5 , \text{cm}} = 15 , \text{cm}^2 \]
Now, since the base of the pyramid is an equilateral triangle, the area \(B\) of an equilateral triangle with side length \(s\) is given by the formula:
\[ B = \frac{\sqrt{3}}{4} s^2 \]
Setting this equal to the base area we found:
\[ \frac{\sqrt{3}}{4} s^2 = 15 \]
To solve for \(s^2\), we multiply both sides by 4:
\[ \sqrt{3} s^2 = 60 \]
Then we divide by \(\sqrt{3}\):
\[ s^2 = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]
Finally, we take the square root to find \(s\):
\[ s = \sqrt{20\sqrt{3}} = \sqrt{20} \cdot \sqrt[4]{3} = 2\sqrt{5} \cdot \sqrt[4]{3} \]
Thus, the side length \(s\) of the equilateral triangle base of the pyramid can be computed, but as it is complex and involves irrational numbers.
In summary, the side length \(s\) of the equilateral triangle base can be approximated or kept in this form for final calculation if precision is needed.