https://file.aiquickdraw.com/mathbot/file/1729829866884vd2u3rt2.jpg

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**.

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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.

The figure shows a container in the shape of an inverted right pyramid VABCD. Its base ABCD is a square of side 16 cm. The container holds 320 cm^3 of water. The water surface PQRS in the container is in the shape of a square of side 8 cm. (a) Find the depth of water in the container. (b) Find the area of the wet surface of the container.

Solve an inverted pyramid volume problem where the container holds 320 cm^3 of water and the base is a square with a side of 16 cm. Learn how to calculate the depth of water and the wet surface area. This geometry problem involves volume, surface area, and Pythagorean theorem.

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