A tower that is part of a hotel has a square base of side 4 metres and a roof in the form of a pyramid - Leaving Cert Mathematics - Question 8(a) - 2011

The roof can be treated as a pyramid with a square base of side length 4 m and height determined previously:

1. Base Area Calculation:

   • The area of the base is: $A_{base} = 4 \times 4 = 16 \text{ m}^2.$

2. Lateral Face Area Calculation (triangular faces):

   • Using the height from (i) to find the triangular face area:
   • Height of triangular face is: $\sqrt{\left(\frac{4}{2}\right)^2 + (12.426)^2} = \sqrt{2^2 + (12.426)^2} = \sqrt{4 + 154.35} = \sqrt{158.35} \approx 12.57.$
   • Thus, height of the triangular face is approximately 12.57 m.
   • Total area of one triangular face: $A_{tri} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 12.57 \approx 25.14 \text{ m}^2.$
   • Since there are 4 triangular faces, total area for lateral faces: $A_{lateral} = 4 \times 25.14 = 100.56 \text{ m}^2.$

3. Total Area of the Roof:

   • Finally, the total area of the roof is: $A_{total} = A_{base} + A_{lateral} = 16 + 100.56 = 116.56 \text{ m}^2.$

To assess the impact of angle errors:

1. Maximum Possible Area Calculation:

   • For maximum area, use angles 47° and 41°:
   • Height Calculation: $\text{Height at } 47° = 12 - 10 \tan(41°) \approx \text{Height}_M = 4.18$
   • The base area's height is still 4; total will be: $\text{Area}_{max} \approx 37 \text{ m}^2.$

2. Minimum Possible Area Calculation:

   • For minimum area, use angles 43° and 45°:
   • Height Calculation: $\text{Height at } 43° = 12 - 10 \tan(45°) \approx \text{Height}_{min} \approx 2.675.$
   • Accordingly, total area: $\text{Area}_{min} \approx 26.72 \text{ m}^2.$

3. Final Range for Area of Roof:

   • Therefore, the overall range for the area is: $26.72 ext{ m}^2 \leq A_{roof} \leq 37.07 ext{ m}^2.$