A toy pyramid (not shown) is made from poly(methyl methacrylate), better known by its trade term Lucite. The toy pyramid has a regular hexagonal base of `15 cm^(2)` and a height of 4 cm. In the base of the pyramid, there is a semispherical indentation 2 cm in diameter. If the pyramid weighs 21.129 g, then what is the density of Lucite? (Density equals mass divided by volume).

Options

A. 1.06 g/`cm^(3)`
B. 1.18 g/`cm^(3)`
C. 2.0 g/`cm^(3)`
D. 6.51 g/`cm^(3)`

Answer

To find the density of the Lucite toy pyramid, we need to calculate the volume of the pyramid and then subtract the volume of the semi-spherical indentation. Finally, we will use the mass of the pyramid to find the density.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Area of the base of the pyramid (A) = 15 cm²
- Height of the pyramid (h) = 4 cm
- Diameter of the semi-spherical indentation = 2 cm
- Mass of the pyramid (m) = 21.129 g

2. **Calculate the Volume of the Pyramid:**
The formula for the volume \( V \) of a pyramid is given by:
\[ V = \frac{1}{3} \times \text{Area of base} \times \text{Height} \]
Substituting the values:
\[ V = \frac{1}{3} \times 15 \, \text{cm}^2 \times 4 \, \text{cm} = \frac{1}{3} \times 60 \, \text{cm}^3 = 20 \, \text{cm}^3 \]

3. **Calculate the Volume of the Semi-Spherical Indentation:**
The formula for the volume \( V_s \) of a sphere is:
\[ V_s = \frac{2}{3} \pi r^3 \]
First, we need to find the radius \( r \):
\[ r = \frac{\text{Diameter}}{2} = \frac{2 \, \text{cm}}{2} = 1 \, \text{cm} \]
Now substituting the radius into the volume formula:
\[ V_s = \frac{2}{3} \pi (1 \, \text{cm})^3 = \frac{2}{3} \pi \approx \frac{2 \times 3.14}{3} \approx 2.094 \, \text{cm}^3 \]

4. **Calculate the Effective Volume of the Toy Pyramid:**
The effective volume \( V_e \) of the toy pyramid is:
\[ V_e = V - V_s = 20 \, \text{cm}^3 - 2.094 \, \text{cm}^3 \approx 17.906 \, \text{cm}^3 \]

5. **Calculate the Density of Lucite:**
The density \( \rho \) is given by the formula:
\[ \rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m}{V_e} \]
Substituting the values:
\[ \rho = \frac{21.129 \, \text{g}}{17.906 \, \text{cm}^3} \approx 1.18 \, \text{g/cm}^3 \]

### Final Answer:
The density of Lucite is approximately \( 1.18 \, \text{g/cm}^3 \).

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