Problem
Part B (9 marks) MS11-4 Performs calculations in relation to two-dimensional and three-dimensional figures. Question 2 A wedding cake with three tiers rests on the table. Each tier is $6 \mathrm{~cm}$ high. The layers have radii of $20 \mathrm{~cm}, 15 \mathrm{~cm}$ and $10 \mathrm{~cm}$ respectively. $2 \pi r \times h$ a) Calculate the area of the cake, that is iced, correct to the nearest $\mathrm{cm}^{2}$ ? (4 marks) \[ \begin{aligned} A^{\prime} & =2 r \times h \\ & =2 \times r \times 10 \times 6 \\ & =376.99 \\ A^{2} & =2 \pi \times \times h \\ & =2 \times \pi \times 15 \times 6 \\ & =565.4 \end{aligned} \] \[ A^{3}=2 \pi r \times h \] b) Find the total volume of the cake, correct to the nearest $\mathrm{cm}^{3}$ ? (2 marks) 6
Solution
Step 1 ：\(A_1 = 2 \pi r_1 h = 2 \pi (20)(6) = 753.98 \text{ cm}^2\)
Step 2 ：\(A_2 = 2 \pi r_2 h = 2 \pi (15)(6) = 565.49 \text{ cm}^2\)
Step 3 ：\(A_3 = 2 \pi r_3 h = 2 \pi (10)(6) = 376.99 \text{ cm}^2\)
Step 4 ：\(A_{total} = A_1 + A_2 + A_3 = 753.98 + 565.49 + 376.99 = 1696.46 \text{ cm}^2\)
Step 5 ：\(V_1 = \pi r_1^2 h = \pi (20)^2 (6) = 7539.82 \text{ cm}^3\)
Step 6 ：\(V_2 = \pi r_2^2 h = \pi (15)^2 (6) = 4241.15 \text{ cm}^3\)
Step 7 ：\(V_3 = \pi r_3^2 h = \pi (10)^2 (6) = 1884.96 \text{ cm}^3\)
Step 8 ：\(V_{total} = V_1 + V_2 + V_3 = 7539.82 + 4241.15 + 1884.96 = 13665.93 \text{ cm}^3\)
Step 9 ：\boxed{\text{Total iced area} \approx 1696 \text{ cm}^2, \text{Total volume} \approx 13666 \text{ cm}^3}
