The problem in the image presents a container in the shape of an inverted right pyramid, where the base $ABCD$ is a square with a side of 16 cm. The container holds 320 cm$^3$ of water, and the water surface forms a square $PQRS$ with a side of 8 cm.

Part (a): Finding the depth of water in the container

We are given:

• The volume of water $V = 320 \, \text{cm}^3$,
• The side length of the water surface $s = 8 \, \text{cm}$,
• The side length of the pyramid's base $S = 16 \, \text{cm}$.

The volume of a pyramid is given by the formula: $V = \frac{1}{3} A_{base} h$ where:

• $A_{base}$ is the area of the square base,
• $h$ is the height (depth) of the water.

For the surface $PQRS$, we use: $A_{base} = s^2 = 8^2 = 64 \, \text{cm}^2$

Substitute into the volume formula: $320 = \frac{1}{3} \times 64 \times h$

Solving for $h$: $320 = \frac{64}{3} h \quad \Rightarrow \quad h = \frac{320 \times 3}{64} = 15 \, \text{cm}$

Thus, the depth of the water in the container is 15 cm.

Part (b): Finding the area of the wet surface of the container

The wet surface consists of:

• The square base $PQRS$,
• The four triangular faces around the water.

The area of the square $PQRS$ is: $A_{PQRS} = s^2 = 8^2 = 64 \, \text{cm}^2$

Next, we need the area of the triangular faces. Each triangular face has a base of 8 cm (side of $PQRS$) and slant height, which can be found using the Pythagorean theorem.

The height of the pyramid from $V$ to the base $ABCD$ is given by the slant height calculation. The slant height for the triangular face is: $\text{Slant height} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \, \text{cm}$

Now, the area of one triangular face is: $A_{triangle} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 8 \times 17 = 68 \, \text{cm}^2$

Since there are 4 triangular faces, the total area of the triangular faces is: $A_{triangles} = 4 \times 68 = 272 \, \text{cm}^2$

Thus, the total wet surface area is: $A_{\text{total}} = A_{PQRS} + A_{triangles} = 64 + 272 = 336 \, \text{cm}^2$

Therefore, the area of the wet surface is 336 cm².

Do you have any questions, or would you like further clarification?

Here are 5 related questions to expand on this problem:

1. How would the formula change if the pyramid were not inverted?
2. What is the formula for the volume of a truncated pyramid?
3. How would the slant height change if the water level was deeper?
4. Can the volume of the water be recalculated for a different base size?
5. How would you calculate the surface area of a regular, non-inverted pyramid?

Tip: Always double-check units when solving geometry problems involving volume and surface area to ensure consistency.