Let's break down the problem step by step.

(a) Calculate the height, $h$, of the container.

We are given:

• Length $l = 15 \, \text{cm}$,
• Width $w = 8 \, \text{cm}$,
• Volume $V = 720 \, \text{cm}^3$.

The volume $V$ of a cuboid is calculated by the formula: $V = l \times w \times h$

Substitute the given values: $720 = 15 \times 8 \times h$ $720 = 120 \times h$

Solve for $h$: $h = \frac{720}{120} = 6 \, \text{cm}$

(b) Calculate the surface area of the outside of the container.

The container has an open top, so we need to calculate the surface area of the sides and the base.

The surface area $A$ of a cuboid with an open top is given by: $A = l \times w + 2 \times (l \times h) + 2 \times (w \times h)$

Substitute the known values: $A = (15 \times 8) + 2 \times (15 \times 6) + 2 \times (8 \times 6)$ $A = 120 + 2 \times 90 + 2 \times 48$ $A = 120 + 180 + 96 = 396 \, \text{cm}^2$

(c) Calculate the height of the liquid in the container.

The liquid fills 60% of the container's volume. Therefore, the volume of the liquid is: $V_{\text{liquid}} = 0.6 \times 720 = 432 \, \text{cm}^3$

Now, using the volume formula for a cuboid, the height of the liquid $h_{\text{liquid}}$ is: $V_{\text{liquid}} = l \times w \times h_{\text{liquid}}$ $432 = 15 \times 8 \times h_{\text{liquid}}$ $432 = 120 \times h_{\text{liquid}}$ $h_{\text{liquid}} = \frac{432}{120} = 3.6 \, \text{cm}$

Summary of Answers:

• (a) Height of the container: $6 \, \text{cm}$
• (b) Surface area of the outside of the container: $396 \, \text{cm}^2$
• (c) Height of the liquid: $3.6 \, \text{cm}$

Would you like any further details or have any questions?

Here are 5 related questions you might consider:

1. What would the surface area be if the container had a lid?
2. How would the surface area change if the height were doubled?
3. How do you calculate the diagonal length of a cuboid?
4. If the container were filled to 80%, what would be the height of the liquid?
5. How do you derive the surface area formula for a cuboid?

Tip: Always recheck your formulas to ensure they're appropriate for the shape and conditions given in the problem (e.g., open vs. closed surfaces).