To find the side length of the equilateral triangle that forms the base of the triangular pyramid, we first need to calculate the volume of the pyramid using its mass and density.
The formula for density is:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
From this, we can rearrange to find the volume:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{400 , \text{g}}{16 , \text{g/cm}^3} = 25 , \text{cm}^3 \]
Next, we know the volume \( V \) of a triangular pyramid (or tetrahedron) is given by:
\[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \]
Let \( A \) be the area of the base (the equilateral triangle). The height of the pyramid is given as \( 5 , \text{cm} \), so:
\[ 25 , \text{cm}^3 = \frac{1}{3} \times A \times 5 , \text{cm} \]
Solving for \( A \):
\[ 25 = \frac{5}{3} A \implies A = 25 \times \frac{3}{5} = 15 , \text{cm}^2 \]
Next, we calculate the area of the base equilateral triangle using the formula:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
Where \( s \) is the side length of the triangle. Setting the equation for area equal to \( 15 \):
\[ \frac{\sqrt{3}}{4} s^2 = 15 \]
Multiplying both sides by \( 4 \):
\[ \sqrt{3} s^2 = 60 \]
Dividing both sides by \( \sqrt{3} \):
\[ s^2 = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]
Now taking the square root of both sides gives us:
\[ s = \sqrt{20\sqrt{3}} = \sqrt{20} \times \sqrt[4]{3} \]
Calculating \( \sqrt{20} \approx 4.47 \) and \( \sqrt[4]{3} \approx 1.316 \), we find:
\[ s \approx 4.47 \times 1.316 \approx 5.89 \text{ cm} \]
Thus, the base side length of the triangular pyramid is approximately 5.89 cm.
The correct answer is 5.89 cm.