The problem consists of two parts related to a composite solid made of a pyramid and a cuboid:

Given Data:

• Pyramid:
  • Height = 28 cm
  • Square base of side = 30 cm
• Cuboid:
  • Square base of side = 30 cm
  • Height = 40 cm

(i) Volume of the Solid

The volume of the solid is the sum of the volumes of the cuboid and the pyramid.

1. Volume of the cuboid (V_cuboid): $V_{\text{cuboid}} = \text{Base Area} \times \text{Height} = (30 \, \text{cm})^2 \times 40 \, \text{cm}$ $V_{\text{cuboid}} = 900 \times 40 = 36,000 \, \text{cm}^3$

2. Volume of the pyramid (V_pyramid): $V_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times (30 \, \text{cm})^2 \times 28 \, \text{cm}$ $V_{\text{pyramid}} = \frac{1}{3} \times 900 \times 28 = 8,400 \, \text{cm}^3$

3. Total Volume (V_total): $V_{\text{total}} = V_{\text{cuboid}} + V_{\text{pyramid}} = 36,000 + 8,400 = 44,400 \, \text{cm}^3$

(ii) Total Surface Area of the Solid

To find the total surface area, we need to account for the outer surface areas of both the cuboid and the pyramid.

1. Surface area of the cuboid (excluding the top face):

   • The total surface area of the cuboid is: $\text{Surface Area of Cuboid} = 2(\text{length} \times \text{width} + \text{length} \times \text{height} + \text{width} \times \text{height})$ Since the base is square: $\text{SA}_{\text{cuboid}} = 2(30 \times 30 + 30 \times 40 + 30 \times 40) = 2(900 + 1200 + 1200) = 2 \times 3300 = 6600 \, \text{cm}^2$
   • The top face (30 cm × 30 cm) is shared with the base of the pyramid, so we exclude that. Its area is: $30 \times 30 = 900 \, \text{cm}^2$
   • So, the total surface area of the cuboid (excluding the top) is: $6600 - 900 = 5700 \, \text{cm}^2$

2. Surface area of the pyramid: The surface area of the pyramid consists of four triangular faces. The area of each triangular face is: $A_{\text{triangle}} = \frac{1}{2} \times \text{Base} \times \text{Slant height}$ To find the slant height $l$, use the Pythagorean theorem: $l = \sqrt{\left(\frac{\text{Base}}{2}\right)^2 + \text{Height}^2} = \sqrt{15^2 + 28^2} = \sqrt{225 + 784} = \sqrt{1009} \approx 31.76 \, \text{cm}$ Now, the area of each triangular face: $A_{\text{triangle}} = \frac{1}{2} \times 30 \times 31.76 \approx 476.4 \, \text{cm}^2$ Since there are four triangular faces: $\text{Total area of pyramid} = 4 \times 476.4 \approx 1,905.6 \, \text{cm}^2$

3. Total surface area: $\text{Total Surface Area} = 5700 + 1905.6 = 7605.6 \, \text{cm}^2$

Final Answers:

(i) Volume of the solid = 44,400 cm³

(ii) Total surface area of the solid = 7605.6 cm²

Would you like more details on any part of the solution? Here are some related questions:

1. How do we find the surface area of complex solids?
2. Can the surface area of the pyramid be computed differently?
3. What happens if the base of the pyramid isn’t square?
4. How would the surface area change if the dimensions of the cuboid changed?
5. Can we derive the slant height for pyramids with different shapes?

Tip: Always make sure to subtract shared or hidden faces when calculating surface areas of composite shapes!